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TECHNICAL  MECHANICS  ""■• 


STATICS    AND    DYNAMICS 


BY 


EDWARD    R.   MAURER 

Professor  of  Mechanics  in  the  University  of  Wisconsin 


FOURTH  EDITION,   REVISED   AND   ENLARGED 
TOTAL  ISSUE  TWENTY-TWO  THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,  Inc. 
London:   CHAPMAN   &  HALL,   Limited 


-   J  ,. 


111? 


Engineering 
Library 


Copyright,  igo3,  1914,  193^ 

BV 

Edward  R.  Maurer 


Stanbope  ipress 

,  GILSON    COMP. 
BOSTON,  U.S.A. 


H.  GILSON    COMPANY  19-01 


PREFACE 


The  following  paragraph  is  an  adaptation  from  the  preface  of  the  first 
edition  of  this  work,  published  ten  years  ago;  it  applies  to  the  present  edition. 
This  book  might  be  described  fairly  as  a  theoretical  mechanics  for  students 
of  engineering.  It  is  not  comparable  to  books  commonly  called  Theoretical 
Mechanics,  generally  intended  for  students  of  mathematics  or  physics;  nor 
to  books  commonly  titled  AppUed  IMechanics  which  generally  include  a  treat- 
ment of  strength  of  materials,  hydraulics,  etc.,  for  students  of  engineering. 
The  title  Technical  Mechanics  seems  fairly  appropriate  for  this  book;  and 
inasmuch  as  it  is  not  otherwise  used  in  this  country,  it  was  so  adopted.  On 
the  theoretical  side,  practically  each  subject  discussed  herein  has  a  direct 
bearing  on  some  engineering  problem.  The  applications  were  selected  and 
presented  for  the  purpose  of  illustrating  a  principle  of  mechanics  and  for 
training  students  in  the  use  of  such  principles,  —  not  to  furnish  information, 
except  incidentally,  about  the  structure,  machine,  or  what  not  to  which  the 
application  was  made. 

Ten  years  use  of  the  book  as  a  text  in  the  author's  classes  has  suggested 
many  changes;  and  in  recent  years  the  need  of  a  new  collection  of  problems 
has  become  urgent.  Accordingly,  a  revision  was  undertaken,  and  the  effort  has 
resulted  in  a  practically  rewritten  book.  Indeed  the  only  portion  of  the  former 
edition  used  again  with  little  or  no  change  is  the  present  Appendix  A.  Though 
containing  fewer  pages  than  the  old  book,  the  new  one  —  because  of  its  (nearly 
one-third)  larger  printed  page  —  contains  more  material  than  the  old. 

Inasmuch  as  Mechanics  deals  mainly  with  subjects  permanent  in  character, 
the  revision  consists  principally  of  changes  in  arrangement  and  presentation. 
Both  were  determined  upon  to  a  large  degree  by  a  desire  to  furnish  an  ade- 
quate course  of  instruction  for  students  in  engineering  in  one  semester,  "five 
times  per  week."  To  this  end,  it  was  necessary  to  sacrifice  logical  order  of 
arrangement  more  or  less.  As  in  former  editions,  Statics  is  presented  first 
because  relatively  simpler  than  Dynamics.  Kinematics,  as  such,  is  not  given 
a  place.  The  chapter  on  Attraction  and  Stress  has  not  been  retained.  Dis- 
cussion of  Friction  and  Efficiency  has  been  amplified,  and  Dynamics  has  been 
extended  to  provide  a  quantitative  explanation  of  simple  gyroscopic  action. 
Many  solved  numerical  examples  have  been  added  to  elucidate  principles. 
The  collection  of  problems  to  be  solved  by  students  has  been  completely 
changed. 


lU 


4942^^5 


IV 


All  of  Statics  except  Arts.  23,  25,  26,  and  27  may  be  mastered  with  no  knowl- 
edge of  mathematics  beyond  trigonometry.  Calculus  methods  are  used 
in  Dynamics,  but  a  good  knowledge  of  the  elements  only  of  that  branch  of 
mathematics  is  presupposed.  Graphical  methods  are  used  freely,  as  much 
as  the  algebraic  in  Statics. 

The  author  is  pleased  to  acknowledge  with  thanks  the  helpful  suggestions 
and  criticisms  of  the  teaching  staff  in  Mechanics  at  the  University  of  Ilhnois; 
of  his  colleague,  Professor  M.  O.  Withey;  and  of  Professor  C.  H.  Burnside 
of  Columbia  University.  He  thanks  also  American  Machinist,  Engineering 
Record,  and  Engineering  News  for  permission  to  copy  and  for  gifts  of  cuts; 
and  individuals  and  other  journals  named  in  the  text  for  similar  favors. 

Madison,  Wisconsin. 
December,  1913. 

To  the  edition  above  described  there  has  been  added  a  second  collection  of 
problems,  pages  354-377;  and  articles  38,  44,  49,  5^,  5^,  55,  5^,  58  have  been 
modified. 

September,  191 7. 


TABLE   OF   CONTENTS 


CHAPTER  I 

COMPOSITION  AND  RESOLUTION  OF  FORCES 

Article  Page 

1.  Introduction i 

2.  Force;  Definitions 4 

3.  Parallelogram  and  Triangle  of  Forces 7 

4.  Composition  of  Concurrent  Forces 11 

5.  Moment  of  a  Force;  Couples 16 

6.  Graphical  Composition  of  Coplanar  Nonconcurrent  Forces 20 

7.  Algebraic  Composition  of  Coplanar  Nonconcurrent  Forces 23 

8.  Moment  of  a  Force;  Couples 27 

9.  Noncoplanar  Nonconcurrent  Forces 3° 

CHAPTER  II 
FORCES  IN  EQUILIBRIUM 

10.  Principles  of  Equilibrium - 34 

11.  Coplanar  Concurrent  Forces 4° 

12.  Coplanar  Parallel  Forces 44 

13.  Coplanar  Nonconcurrent  Nonparallel  Forces 46 

14.  Noncoplanar  Forces 5° 

CHAPTER  III 
SIMPLE  STRUCTURES 

15.  Simple  Frameworks  (Truss  Type) 54 

16.  Graphical  Analysis  of  Trusses;  Stress  Diagrams 59 

17.  Simple  Frameworks  (Crane  Type) 64 

18.  Cranes 69 

CHAPTER  IV 
FRICTION 

19.  Definitions  and  General  Principles 74 

20.  Friction  in  Some  Mechanical  Devices 78 

CHAPTER  V 
CENTER  OF  GRAVITY 

Center  of  Gravity  of  Bodies 86 

Centroids  of  Lines,  Surfaces,  and  Solids 9° 

23.  Centroids  Determined  by  Integration 93 

24.  Centroids  of  Some  Lines,  Surfaces,  and  Solids 98 

V 


21. 
22. 


VI 


CHAPTER  VI 


SUSPENDED  CABLES,  WIRES,  CHAINS,  ETC. 

Article  Page 

25.  Parabolic  Cable 102 

26.  Catenary  Cable 107 

27.  Cable  with  Concentrated  Loads 113 


CHAPTER  VII 
RECTILINEAR  MOTION 

28.  Velocity  and  Acceleration 118 

29.  Motion  Graphs 126 

30.  Simple  Harmonic  Motion 131 

31.  Motion  and  Force 138 


CHAPTER  VIII 
CURVILINEAR  MOTION 

32.  Velocity  and  Acceleration 144 

33.  Components  of  Velocity  and  Acceleration 148 

34.  Motion  of  the  Center  of  Gravity  of  a  Body 155 


CHAPTER  IX 
TRANSLATION  AND  ROTATION 

35.  Translation 163 

36.  Moment  of  Inertia  and  Radius  of  Gyration 168 

37.  Rotation 176 

38.  Axle  Reactions 180 

39.  Pendulums 183 


CHAPTER  X 
WORK,  ENERGY,  POWER 

40.  Work 189 

41.  Energy 193 

42.  Power 196 

43.  Principles  of  Work  and  Energy 203 

44.  Efficiency;  Hoists 211 

45.  Kinetic  Friction 221 


CHAPTER  XI 
MOMENTUM  AND  IMPULSE 

46.  Linear  Momentum  and  Impulse 228 

47.  Impact  or  Collision 232 

48.  Angular  Momentum  and  Impulse 237 

49.  Gyrostat 243 


vu 

CHAPTER  XII 

TWO  DIMENSIONAL  (PLANE)  MOTION 

Article  Page 

50.  Kinematics  of  Plane  Motion 256 

51.  Kinetics  of  Plane  Motion 261 

52.  Rolling  Resistance 268 

53.  Relative  Motion 273 


CHAPTER  XIII 
THREE  DIMENSIONAL  (SOLID)  MOTION 

54.  Body  With  a  Fixed  Point,  Kinematics  of 280 

55.  Body  With  a  Fixed  Point,  Kinetics  of 284 

56.  Gyrostat 288 

57.  Principal  Moments  of  Inertia,  and  Axes 292 

58.  Any  Solid  Motion;  Summary  of  Dynamics 296 

APPENDIX   A.     THEORY   OF  DIMENSIONS   OF  UNITS 302 

APPENDIX   B.     MOMENT  OF  INERTIA  OF  PLANE  AREAS 308 

PROBLEMS 323 


TECHNICAL  MECHANICS 


I.   Introduction 

Mechanics  had  its  origin  in  the  experience  of  ancient  peoples  with  de- 
vices for  lifting  and  moving  heavy  things.  The  devices  included  the 
so-called  simple  machines  or  mechanical  powers;  namely,  the  lever,  the 
pulley,  the  wheel  and  axle,  the  inclined  plane,  the  wedge  and  the  screw. 
That  experience  probably  afforded  fairly  definite  and  full  knowledge  of  the 
practical  advantages  of  these  various  devices,  but  the  simple  and  precise 
mechanical  principles  involved  in  them  were  long  unrecognized.  The  first 
recognition  of  such  a  principle  marked  the  real  beginning  of  the  science  of 
Mechanics. 

History  records  that  the  principle  of  the  lever  is  the  mechanical  principle 
first  discovered,  and  that  Archimedes  (287-212  B.C.),  famous  Greek  mathe- 
matician, was  the  discoverer.  He  perceived  the  application  of  this  prin- 
ciple to  the  wheel  and  axle  (continuous  lever),  to  the  pulley  (movable 
lever),  and  to  certain  combinations  or  systems  of  pulleys  and  cords,  one  of 
which  still  bears  his  name.  The  discovery  of  the  principle  of  buoyant  effort 
on  a  body  floating  on  or  immersed  in  a  fluid  is  due  to  him.  Apparently  no 
additions  to  these  achievements  of  Archimedes  were  made  during  the  sixteen 
centuries  following  his  time. 

The  principle  of  the  lever  as  understood  by  Archimedes  covered  only  the 
special  case  of  two  heavy  weights  suspended  from  a  horizontal  bar  sup- 
ported at  a  point  (fulcrum)  between  them.  For  such  case  he  stated  that 
the  weights  are  inversely  as  the  distances  from  the  fulcrum  to  the  points  of 
suspension.  The  principle  was  extended  to  include  the  case  of  forces  ap- 
plied obliquely,  by  Leonardo  da  Vinci  (145  2-1 5 19),  famous  Italian  artist  and 
engineer.  He  perceived  that  the  efficacy  of  such  a  force  depends  on  the 
distance  from  the  fulcrum,  not  to  the  point  of  application  of  the  force,  but 
to  its  line  of  action. 

The  principle  next  discovered  was  that  of  the  inclined  plane,  first  defi- 
nitely stated  by  Simon  Stevin  (i 548-1620),  Dutch  mathematician  and  en- 
gineer. His  statement  of  the  principle  was  somewhat  as  follows:  The  force 
(acting  along  the  plane)  required  to  support  a  (frictionless)  body  resting 
upon  it  is  to  the  weight  of  the  body  as  the  height  of  the  plane  is  to  its 
length  (measured  along  the  slope).  This  principle  afforded  the  explana- 
tion of  the  wedge  (double  inclined  plane)  and  the  screw  (continuous  inclined 
plane).      Stevin  deduced   the   parallelogram  law  for    two    forces    at  right 


2     '  ■  '  Art.  I 

•a'n^ies'  'from  the  principle  of  the  inclined  plane;  and  from  his  study  of 
pulleys  he  noted  that  what  is  gained  in  power  is  lost  in  speed.  Thus  he 
caught  the  first  glimpse  of  two  important  principles,  —  that  of  the  parallelo- 
gram of  forces,  and  that  of  virtual  velocity  or  work. 

The  first  discoveries  of  laws  of  motion  were  made  by  Galileo  (i  564-1 642), 
Italian  astronomer  and  physicist.  For  2000  years  it  had  been  believed  that 
heavy  bodies  fall  more  rapidly  than  light  ones.  This  Galileo  disproved  by 
actual  trial  at  the  leaning  tower  of  Pisa.  Next  he  was  led  to  inquire  about 
the  manner  in  which  a  body  falls,  or  how  the  speed  changes.  He  made 
several  guesses  at  this  law,  and  finally  verified  one  of  them  by  indirect 
experiment  and  deduction.  Up  to  Galileo's  time,  it  was  believed  that  rest 
was  the  natural  condition  for  a  body;  and  that  motion  was  unnatural, 
requiring  some  outside  cause  (force)  to  maintain  it,  and  ceasing  only  when 
the  force  ceases.  Galileo  perceived  that  motion  is  just  as  natural  as  rest; 
that  motions  cease  not  because  they  are  unnatural,  but  because  of  some 
influence  (force)  from  the  outside  operating  to  reduce  the  motion  and 
eventually  to  destroy  it.  In  short,  he  discovered  the  so-called  first  law  of 
motion,  usually  credited  to  Newton.     He  invented  the  telescope. 

Huygens  (1629-1695),  Dutch  physicist,  made  some  important  contribu- 
tions to  this  science.  He  developed  the  theory  of  the  pendulum,  determined 
the  acceleration  due  to  gravity  from  pendulum  obser\^ations,  and  deduced 
certain  theorems  regarding  centrifugal  force.  He  invented  the  clock  pen- 
dulimi  and  escapement. 

Newton  (1642-1727),  English  mathematician  and  physicist,  is  generally 
regarded  as  the  founder  of  Mechanics.  At  an  early  age  he  began  an  at- 
tempt to  explain  the  motions  of  the  planets,  whose  orbits  and  speeds  were 
then  well  known,  in  terms  of  experience  with  more  familiar  motions.  He 
succeeded  in  thus  explaining  many  features  of  the  planetary  motions,  and 
established  that  there  are  certain  principles  common  to  the  motion  of  all 
bodies,  celestial  and  terrestial.  These  principles  are  generally  known  as 
Newton's  laws  of  motions  (see  index).  His  study  of  planetary  motion  led 
to  other  great  achievements,  among  which  may  be  mentioned  the  discovery 
of  the  law  of  universal  gravitation,  and  the  invention  of  the  calculus  (also 
invented  independently  by  Leibnitz,  German  mathematician). 

Since  Newton,  "no  essentially  new  principle  [of  Mechanics]  has  been 
stated.  All  that  has  been  accomplished  since  his  day  has  been  a  deductive, 
formal,  and  mathematical  development  on  the  basis  of  Newton's  laws."* 
Such  development  consritutes  the  body  of  knowledge  which  we  call  Me- 
chanics, or  sometimes  Rational  and  Theoretical  Mechanics,  to  distinguish  it 
from  Applied  Mechanics.  It  may  be  defined  as  the  science  of  motion,  but  it 
includes  the  science  of  rest  as  a  relatively  minor  part. 

*  For  a  full  and  critical  account  of  that  development,  see  Mach's  "  Science  of  Me- 
chanics," from  which  the  quotation  was  taken,  or  Co.x's  "  Mechanics  "  for  a  good  but  less 
critical  account. 


Art.  I  « 

Adaptations  of  rational  mechanics  have  played  an  important  part  in 
the  development  of  the  science  of  engineering,  particularly  in  the  depart- 
ments of  structures  and  machines.  Such  adaptations,  together  with  our 
knowledge  of  friction,  strength  of  materials,  and  certain  properties  of  fluids, 
constitute  Applied  Mechanics.  Among  the  pioneer  workers  in  this  field 
should  be  mentioned  the  following:  Coulomb  (1736-1806),  Navier  (1785- 
1836),  Poncelet  (1788-1867),  Morin  (1795-1880),  Saint-Venant  (1797-1886), 
Weisbach  (1806-71),  Rankine  (1820-72),  Grashof  (1826-93)  and  Bauschinger 
(1834-93).* 

Under  Technical  Mechanics,  the  present  author  includes  those  prin- 
ciples of  rational  mechanics  which  are  especially  applicable  in  various 
fields  of  engineering,  and  some  of  our  knowledge  of  friction.  The  book  is 
divided  into  two  parts  called  Statics  and  Dynamics.  The  first  deals  with 
certain  of  the  circumstances  of  bodies  at  rest,  and  the  second  with  those  of 
bodies  in  motion.  The  certain  circumstances  dealt  with  will  become  ap- 
parent to  the  student  as  he  progresses  in  the  subject. 

*  See  Keek's  "  Mechanik  "  for  an  account  of  their  work  and  fuller  list. 


STATICS 


CHAPTER   I 

COMPOSITION  AND   RESOLUTION   OF  FORCES 
2.  Force;  Definitions 

Bodies  act  upon  each  other  in  various  ways,  producing  different  kinds 
of  results.  Any  action  of  one  body  upon  another  which,  when  exerted 
alone,  would  result  in  motion  of  the  body  acted  upon,  or  in  change  of  motion 
if  the  body  is  already  moving,  is  called  force;  the  word  is  a  general  term  for 
push  and  pull.  Our  earliest  notions  about  forces  are  based  on  our  experience 
with  forces  exerted  by  or  upon  ourselves.  Through  this  experience  we  have 
learned  that  a  force  has  magnitude,  place  of  application,  and  direction, 
sometimes  called  the  characteristics  of  a  force. 

To  express  the  magnitude  of  a  force,  we  must  of  course  compare  it  to 
some  other  force  regarded  as  a  unit.  Many  units  of  force  are  in  use;  the 
most  convenient  are  the  so-called  gravitation  units.  They  are  the  earth- 
pulls  on  our  standards  for  measuring  quantity  of  material  (as  iron,  coal, 
grain,  sugar,  etc.),  commonly  called  standards  of  weight.*  The  earth-pull 
on  any  of  these  standards  is  called  by  the  name  of  the  standard;  thus  the 
earth-pull  on  the  pound  standard  (also  any  equal  force)  is  called  a  pound; 
the  earth-pull  on  the  kilogram  standard  (also  any  equal  force)  is  called  a 
kilogram,  etc.  Since  the  earth-pull  on  any  given  thing  varies  in  amount 
as  the  thing  is  transported  from  place  to  place,  gravitation  units  of  force 
are  not  constant  with  regard  to  place.  But  this  variation  need  not  be 
regarded  in  most  engineering  calculations  because  any  error  due  to  such 
disregard  is  generally  smaller  than  errors  due  to  other  approximations  in 
the  calculations.  The  extreme  variation  in  any  gravitation  unit  is  that 
between  its  magnitudes  at  the  highest  elevation  on  the  equator  and  at  the 
poles;  this  difference  is  but  0.6  per  cent.  For  points  within  the  United 
States   the   extreme  variation   equals   about   0.3    per   cent.     For   any  two 

*  In  common  parlance  the  word  weight  is  used  in  at  least  two  senses.  Thus,  suppose 
that  a  dealer  sells  coal  to  a  consumer  by  weight,  and  engages  a  teamster  to  deliver  it  by 
weight;  to  the  consumer,  the  weight  of  each  wagon  load  represents  a  certain  amount  of 
useful  material,  but  to  the  teamster  it  represents  a  certain  burden  on  his  team  due  to  the 
action  of  gravity  on  the  coal.  That  is,  weight  suggests  material  to  the  one  man  and 
earth-pull  to  the  other. 


Art.  2  5 

points  on  the  surface  of  the  earth,  the  variation  equals  that  in  the  values  of 
g  in  the  formula 

g  =  32.0894  (i  +  0.0052375  sin^O  (i  ~  0.0000000957  e) 

computed  for  the  two  places;  I  denotes  latitude,  and  e  elevation  above  sea 
level,  in  feet. 

The  place  of  application  of  most  forces  with  which  we  shall  deal  is  a 
portion  of  the  surface  of  the  body  to  which  the  force  is  applied.  A  notable 
exception  is  earth-pull,  or  gravity,  which  is  applied  not  to  the  surface  of  a 
body  but  throughout  the  same.  All  such  are  called  distributed  forces.  The 
places  of  application  of  some  forces  are  very  small  compared  to  the  sur- 
faces of  the  bodies  to  which  they  are  applied,  and  for  many  purposes  these 
places  may  be  regarded  as  points  of  application;  any  such  force  is  called  a 
concentrated  force.  The  line  of  action  of  a  concentrated  force  is  a  line 
indefinite  in  length,  parallel  to  the  direction  of  the  force,  and  containing  its 
point  of  application.  A  concentrated  force  may  act  along  its  line  of  action  in 
one  of  two  ways, — to  the  right  or  left,  up  or  down,  etc.  We  say  that  the  sense 
of  a  force  is  toward  the  right,  toward  the  left,  up,  or  down  as  the  case  may 
be.     That  is,  sense  refers  to  "  arrow-headedness  "  (see  next  paragraph). 

Since  a  force  is  a  vector  quantity,*  it  can  be  represented  in  part  by  a 
vector  (a  straight  line  of  definite  length  and  direction),  the  length  of  the 


a 


vector  representing  the  magnitude  of  the  force  according  to  some  scale, 
and  the  direction  of  the  vector  giving  the  direction  of  the  force.  Thus,  if 
the  pressures  of  the  driving  wheels  of  the  locomotive  on  the  rails  (Fig.  i) 
is  12  tons,  then  the  vector  Aa  (0.4  inch  long)  represents  the  magnitude  and 
direction  of  the  pressures,  the  scale  being  one  inch  "  equals  "  30  tons.  If 
the  force  to  be  represented  is  a  concentrated  one,  as  in  the  illustration,  then 
the  line  of  action  also  can  be  represented  by  the  same  vector  which  repre- 
sents the  force  magnitude  by  drawing  it  through  the  point  of  application 
of  the  force.  Thus  the  vector  Bb  represents  magnitude,  line  of  action,  and 
direction  of  the  pressure  of  the  first  driving  wheel.  We  might  extend  this 
scheme  further  so  as  to  indicate  also  point  of  application  of  the  force  by 
the  head,  say,  of  the  vector  as  Cc;  but  we  will  not  plan  to  do  that  because 
the  point  of  application  is  not  of  importance  in  this  subject,  —  Statics. 

*  A  vector  quantity  is  one  having  magnitude  and  direction,  as,  for  example,  a  definite 
displacement  of  a  moving  point.  A  quantity  having  magnitude  only,  as  the  volume  of  a 
thing  for  example,  is  a  scalar  quantity. 


Chap,  i 


f 


1 


\ 


B 


Fig.  2 


This  unimportance  of  the  point  of  application  is  definitely  expressed  in 
the  principle  of  transmissibility  of  force,  which  for  the  present  purpose  may 

be  stated  as  follows:  The  effect  of  any  force 
applied  to  a  rigid  body  at  rest  is  the  same, 
no  matter  where  in  its  own  line  of  action 
the  force  is  applied.  The  principle  may  be 
roughly  verified  by  experiment,  when  the 
body  on  which  the  force  acts  is  at  rest,  with 
the  apparatus  represented  in  Fig.  2;  it  con- 
sists of  a  rigid  body  suspended  from  two 
spring  balances.  The  springs  are  elongated 
on  account  of  the  weight  of  the  body,  and  if 
a  force,  as  F,  be  applied  at  A,  the  springs  will  suffer  additional  elonga- 
tions which  in  a  way  are  a  measure  of  the  effect  of  the  applied  force. 
If  the  point  of  application  of  F  be  changed  to  B  or  C,  the  spring  readings 
will  not  change;  hence  the  effect  of  F  will  not  have  changed. 

Generally,  when  many  forces  are  to  be  represented  graphically  and  dis- 
cussed, it  would  be  well  to  represent  each  force  by  a  line  and  a  vector,  the 
first  to  represent  the  line  of  action  of  the  force 
and  the  second  to  represent  the  magnitude  and 
the  direction  of  the  force.  Of  course  the  line 
must  be  drawn  through  the  point  of  application 
of  the  force,  but  the  vector  may  be  drawn  where 
convenient.  For  example,  consider  the  forces  act- 
ing on  the  upper  end  of  the  boom  (Fig.  3)  of  a 
derrick.  There  are  three  forces;  namely,  a  down- 
ward force  at  pin  i,  one  toward  the  left  at  pin  2, 
and  one  downward  at  pin  3.  The  lines  marked 
ah,  cd,  and  ef  are  the  lines  of  action  of  the  forces 
respectively;  the  vectors  AB,  CD,  and  EF  (drawn 
where  convenient  but  of  proper  length  and  direc- 
tion) represent  the  magnitudes  and  directions  of 
the  forces.  The  scheme  of  notation  here  used  — 
two  lower-case  letters  on  opposite  sides  of  the  line 
of  action  of  a  force,  and  the  same  capital  letters 
at  the  ends  of  the  vector  representing  its  value  — 
is  in  common  use.  Any  force  so  marked  is  re- 
ferred to  in  written  statement  by  the  two  capitals 
used;  thus  the  first  force  mentioned  above  would 
be  called  the  force  AB.  The  part  of  the  drawing  in  which  the  lines  of  action 
of  the  forces  and  the  body  (here,  a  derrick-boom)  are  represented  is  called  a 
space  diagram;  the  part  in  which  the  vectors  are  drawn  is  called  a  vector 
diagram.  The  scales  of  these  diagrams  are  of  course  different;  the  lengths  of 
lines  in  the  first  represent  distances,  and  those  in  the  second,  force  magnitudes. 


Fig. 


Art.  3  7 

Any  number  of  forces  collectively  considered  is  called  a  system  or  a  set 
of  forces.  The  forces  of  a  set  are  called  coplanar  if  their  lines  of  action  are 
in  the  same  plane,  and  noncoplanar  if  not  in  the  same  plane;  they  are 
called  concurrent  if  their  lines  of  action  intersect  in  a  point,  and  noncon- 
current  when  they  do  not  so  intersect;  they  are  called  parallel  if  their  lines 
of  action  are  parallel,  and  nonpar allel  if  the  lines  of  action  are  not  parallel. 
Force-sets  are  also  described  in  accordance  with  the  foregoing  definitions; 
thus,  a  concurrent  set,  a  noncoplanar  parallel  set,  etc.,  according  as  the 
forces  of  the  set  are  concurrent,  noncoplanar  and  parallel,  etc.  Force- 
sets  can  be  classified  in  various  ways,  as  below  for  example^  — 


Coplanar 


,  fcolinear i 

concurrent         <  ,,  • 

\  nonparallel  ....  2 

nonconcurrent  -!  ^^^^  ^  W  i  '  "  '  ^ 

\nonparallel  ....  4 


concurrent 5 

Noncoplanar    <!  nonconcurrent  /parallel  ._. .  ....     6 


L... ....,     „, 

L  l^nonparallel ....     7 


Two  sets  of  forces  acting  on  a  rigid  body  are  said  to  balance,  when  their 
combined  effect  on  the  rest  or  on  the  motion  of  that  body  is  nil,  so  that 
if  the  body  is  at  rest,  for  example,  then  it  would  remain  at  rest  even  if  all 
the  forces  ceased  to  act.  Two  sets  of  forces  acting  on  a  rigid  body  are  said 
to  be  equivalent  if  either  set  would  balance  the  other  set  reversed  (sense  of 
each  force  changed);  or,  what  amounts  to  the  same  thing,  if  each  set  acting 
singly  would  balance  some  other  third  set.  The  resultant  of  a  set  of  forces 
is  the  single  force  which  is  equivalent  to  the  set;  or,  if  no  single  force  is 
equivalent  to  the  set,  then  the  resultant  is  the  simplest  equivalent  set.  The 
resultant  of  a  set  of  forces  acting  on  a  rigid  body  consists  always  of  a  single 
force  or  of  two  forces  (proved  later).  Having  given  a  set  of  forces,  the  process 
of  finding  a  simpler  equivalent  set  is  called  composition  of  the  given  set.  The 
component  of  a  given  force  is  any  one  of  a  set  which  is  equivalent  to  that 
force.  Having  given  a  force,  the  process  of  finding  a  set  equivalent  to  that 
force  is  called  resolution  of  the  force. 

The  anti-resultant  of  a  set  of  forces  is  the  reversed  resultant  of  the  set.  The 
equilibrant  of  a  set  of  forces  is  the  single  force,  or  pair  of  forces  if  necessary, 
which  could  balance  the  set.  Obviously  the  anti-resultant  and  the  equilibrant 
of  a  set  are  identical. 

3.   Parallelogram  and  Triangle  of  Forces 

The  parallelogram  and  the  triangle  of  forces  are  names  of  certain  methods 
for  determining  (a)  the  resultant  of  two  given  concurrent  forces,  and 
(b)  two  concurrent  components  of  a  given  force. 

§  I.  Composition  of  Two  Concurrent  Forces. — Parallelogram  Law. — 
If  two  forces  acting  upon  a  rigid  body  be  represented  by  lines  OA  and  OB, 
then  their  resultant  is  represented  by  the  diagonal  OC  of  the  parallelogram 


8 


Chap,  i 


OABC.  For  example,  take  the  two  forces  applied  to  the  cap  of  the  boom 
of  Fig.  3  at  points  i  and  2,  their  value  being  2  and  1.2  tons  respectively, 
let  us  suppose.  Extending  the  lines  of  action  to  their  intersection  O 
(Fig.  4),  then  making  OA  =  2  tons  and  OB  =1.2  tons  according  to  some 
convenient  scale,  and  completing  the  parallelogram,  we  get  OC,  and  ac- 
cording to  the  law,  this  line  represents  the  resultant  completely;  that  is, 
the  magnitude  of  the  resultant  is  OC  =^2.2  tons,  the  line  of  action  of 
the  resultant  is  colinear  with  OC,  and  the  sense  of  the  resultant  is  from 
OtoC. 

The  law  can  be  verified  by  means  of  the  apparatus  shown  in  Fig.  5.  It 
consists  of  a  drawing  board  mounted  in  a  vertical  position,  two  pulleys,  a 
spring  balance,  two  weights,  some  cord,  and  a  small  ring.  When  the 
weights  Wi  and  W2  are  suspended  somewhat   as  shown,  then  the  ring  is 


Fig.  4 


Fig.  s 


Fig.  6 


subjected  to  three  forces:  pull  Pi  =  Wi,  pull  P2  =  W2,  and  an  upward 
pull  P3,  the  magnitude  of  which  is  indicated  by  the  spring  balance.  Since 
P3  is  the  equilibrant  of  Pi  and  P2,  the  resultant  of  Pi  and  P2  is  equal  and 
opposite  to  and  colinear  with  P3.  It-  remains  now  to  ascertain  whether  a 
construction  for  the  resultant  of  Pi  and  P2  according  to  the  parallelogram 
law  will  represent  a  force  equal  and  opposite  to  and  colinear  with  Pg,  So 
we  lay  off  OA  and  OB  on  the  board,  just  under  the  strings,  equal  to  Pi 
and  P2,  and  complete  the  parallelogram  OABC;  then  measure  OC  and  com- 
pare its  direction  with  P3.  We  find  that  OC  equals  P3  (by  scale),  and  is 
colinear  with  P3. 

To  test  the  law  for  forces  having  different  points  of  application,  the 
apparatus  shown  in  Fig.  6  might  be  used;  it  consists  of  a  tub  of  water, 
a  floating  drawing  board,  three  smoothly  running  pulleys,  three  weights 
(Wi,  W2,  and  W3),  and  three  cords.  Nails  are  driven  into  the  drawing 
board  at  any  points  Ni,  N2,  and  N3;  the  weights  are  then  suspended  by 
cords  passing  over  the  pulleys,  and  tied  to  the  nails  as  shown;  then  if 
each  weight  is  less  than  the  sum  of  the  other  two,  the  board,  if  not  too 
large,  will  move  about  and  assume  a  position  of  rest  without  touching 
the  tub.  In  such  position,  the  forces  acting  on  the  board  are  its  weight 
(or  gravity),  pressure  of  the  water,  and  the  three  pulls   (Pi,  P2,  and  P») 


Art.  3  g 

practically  equal  to  Wi,  W2,  and  W3  respectively.  Obviously  the  first  two 
forces  balance  each  other;  therefore  the  three  pulls  also  balance,  and  so 
the  resultant  of  Pi  and  P2  is  ec^ual  and  opposite  to  and  colinear  with  P3. 
We  next  determine  the  resultant  R  of  Pi  and  P2  by  the  parallelogram  law: 
extend  the  lines  of  action  of  the  pulls  Pi  and  P2  to  their  intersection  0; 
from  there  lay  off  OA  and  OB  equal  (by  some  convenient  scale)  to  Pi  and 
P2;  complete  the  parallelogram  OABC.  Then  OC  represents  R;  on  compari- 
son it  will  be  found,  as  before,  that  OC  is  equal  and  opposite  to  and  colinear 
with  P3,  and  hence  OC  does  represent  the  magnitude  and  line  of  action  of 
R.  Since  P3,  and  hence  R,  passes  through  0  (the  intersection  of  Pi  and 
P2),  this  experiment  emphasizes  the  fact  that  the  line  of  action  of  the  re- 
sultant of  two  concurrent  forces  passes  through  their  point  of  concurrence. 
The  point  of  application  of  R  might  of  course  be  taken  anywhere  in  OC  or 
its  extension;  for,  so  taken,  R  obviously  would  balance  P3.* 

The  Triangle  Law.  —  If  two  concurrent  forces  acting  on  a  rigid  body  be 
represented  in  magnitude  and  direction  hy  AB  and  BC,  then  their  resultant 
is  represented  in  magnitude  and  direction  by  the  side  ^C  of  the  triangle 

*  By  using  accurate  apparatus  the  foregoing  tests  for  verifying  the  parallelogram  law- 
can  be  made  very  accurately.  Such  verifications  are  as  satisfying  to  many  students  as 
"  mathematical  proof."  What  about  such  proof?  Some  writers  assert  that  the  law  is 
fundamental,  and  not  susceptible  of  deduction  from  anything  more  simple  and  obvious 
than  the  law  itself.  But  many  deductions  or  proofs  have  been  proposed.  All  necessarily 
depend  upon  one  or  more  axioms  or  statements  whose  truth  is  justified  by  experience.  We 
give  a  proof  based  upon  a  principle  of  moments  (Art.  5)  which  most  students  readily 
grant  as  axiomatic  or  justified  by  their  experience.  The  principle  is  that  the  moment  of 
the  resultant  of  two  concurrent  forces  about  any  point  in 
their  plane  equals  the  algebraic  sum  of  the  moments  of  the  E/-,_ 

two  forces  about  the  same  point.     Let  P  and  Q  denote  the  /       ~"'~---,,  ^ 

two  concurrent  forces  and  R  their  resultant.  Suppose  that 
P  and  Q  act  in  OA  and  OB  respectively  (Fig.  7)  —  the 
body  upon  which  they  act  is  not  represented  —  and  let  the  ^ 

lengths  OA   and  OB  represent  the  magnitudes  of  the  forces  / 

P  and  Q  to  some  scale,  that  is  OA  ^  OB  =  P  ^  Q.     OABC  / 

is  a  parallelogram,  and  CD,  CE,  BF,  and  BG  respectively        /    ^ 

are  perpendicular  to  OA,OB,  OA,  and  OC.     Now  the  mo-      /^_ ^_^ 

ments  of  P  and  Q  about  0  equal  zero;  it  follows  from  the  '^0  p        A  D 

principle  of  moments  that  the  moment  of  R  about  0  equals  Pi^    _ 

zero  also,  and  hence  the  line  of  action  of  R  passes  through 

0.  Now  the  area  of  the  parallelogram  is  OA  X  CD;  also  OB  X  CE.  Hence,  OA  -^  OB 
=  CE^  CD;  and  P  -^  Q  =  CE  ^  CD,  or  P  X  CD  =  Q  X  CE;  that  is,  the  moments 
of  P  and  Q  about  C  are  equal.  But  these  two  moments  are  opposite  in  sign,  and  so  their 
algebraic  sum  equals  zero.  It  follows  from  the  principle  of  moments  that  the  moment 
of  R  about  C  equals  zero,  and  hence  the  line  of  action  of  R  passes  through  C.  The  moments 
of  P,  Q,  and  R  about  B  are  respectively,  P  X  BF,  o,  and  R  X  BG;  then,  according  to  the 
principle  of  moments,  R  X  BG  =  P  X  BF,  or  R  ^  P  =  BF  ^  BG.  The  area  of  the  paral- 
lelogram is  OC  X  BG;  also  OA  X  BF.  Hence,  OC  ^  OA  =  BF  ^  BG;  and  from  the  last 
proportions  R  ^  P  =  OC-i-OA;  that  is,  OC  represents/?  according  to  the  same  scale  that 
OA  represents  P. 


lO 


Chap,  i 


Fig.  8 


ABC.  For  example,  let  two  forces  of  2  and  1.2  tons  be  applied  at  i  and  2 
(Fig.  8)  as  shown.  li  AB  and  BC  be  drawn  anywhere  in  the  directions 
of  these  forces,  and  AB  and  BC  be  made  equal  to  the  forces  respectively, 

then  AC  gives  the  magnitude  and  the  direction  of  the 
resultant;  the  line  of  action  of  the  resultant  is  ac,  —  par- 
allel to  ^C  and  concurrent  with  the  given  forces. 

The  resultant  of  two  concurrent  forces  can  be  deter- 
mined without  a  scale  drawing  of  a  triangle  or  parallelo- 
gram. We  sketch  the  triangle  of  forces  roughly,  and  then 
solve  the  triangle  for  the  length  and  direction  of  the  side 
representing  the  resultant.  For  example,  let  the  forces  P 
and  Q  (Fig.  9)  *  equal  100  and  150  pounds  respectively, 
and  the  angle  </>  between  them  be  60  degrees;  required, 
their  resultant  R.  Rouglily,  ABC  is  the  triangle  for  the 
forces,  AC  representing  the  magnitude  and  direction  of 
R,  and  the  angle  ABC  =  180°  —  60°  =  120°.  Then  from 
the  trigonometry  of  the  triangle,  R^  =  100^  +  150^  —  2  X  100  X  150  cos  120° 
=  47,500,  or  R  =  218.3;  2.1so  sin  CAB/sin  120°  =  150/i?,  or  CAB  (the  angle  a 
between  R  and  P)  =  36°  35'.  Employing  the 
foregoing  method,  the  following  general  form- 
ulas may  be  worked  out  for  determining  the 
magnitude  and  direction  of  the  resultant,  — 

Ri  =  p2-\-Qi^  2  PQ  cos  <^; 

sin  a  =  sin  <^  •  Q/R,  and  sin  /3  =  sin  </>  •  P/R, 

where  </>,  a,  and  0  are  the  angles  marked  in  Fig.  9.  When  the  two  forces 
P  and  Q  are  at  right  angles  to  each  other  (0  =  90  degrees) ,  then 

^2  =  p2_|_Q2^     and      tarn  a-=Q/ P. 

§  2.  Resolution  of  a  Force  into  Concurrent  Components  can  be 
accomplished  by  applying  the  triangle  or  parallelogram  law  inversely.  Thus, 
let  it  be  required  to  resolve  the  force  F  (Fig.  10)  into  two  components.  We 
draw  AB  anywhere  equal  (by  some  scale)  and  parallel  to  F;  join  any  point  C 
with  A  and  B,  and  draw  lines  through  any  point  in  ab  parallel  to  ^C  and 
BC;  then  AC  and  CB  represent  the  magnitudes  and  directions,  ac  and  cb 
the  lines  of  action  of  two  forces  equivalent  to  F,  that  is,  components  of  F. 
For  the  resultant  of  these  two  component  forces  is  F,  as  shown  by  the  tri- 

*  For  convenience  and  clearness  of  figure,  a  subdivided  square  (or  rectangle)  will  herein- 
after represent  a  machine,  or  structure  (derrick -boom,  bridge,  etc.),  on  which  the  forces 
under  discussion  act.  If  he  prefers,  the  student  might  regard  the  subdivided  square  as 
representing  a  drawing  board  or  some  other  definite  object  suggested  by  the  square.  It 
is  important  that  he  should  have  in  mind  the  fact  that  forces  act  only  on  material  things 
(bodies),  and  that  the  lines  of  action  of  the  forces  represented  in  any  given  figure  are 
definitely  related  to  th«  body  on  which  the  forces  act. 


Art.  4 


II 


angle  law  applied  directly.  Since  C  was  taken  at  random,  it  is  plain  that  a 
given  force  can  be  resolved  into  many  different  pairs  of  components. 

If  conditions  be  imposed  on  the  components,  the  resolution  is  more  or 
less  definite.  Thus,  let  it  be  required  to  resolve  F  (Fig.  n),  equal  to  350 
pounds,  into  two  components,  one  of  which  must  act  along  the  left-hand  edge 
of  the  board  and  the  other  through  the  lower  right-hand  corner.  Since  the 
three  forces  must  be  concurrent,  the  second  component  must  act  through 
point  i;  so  we  make  AB  equal  and  parallel  to  F  and  draw  from  A  and  B 
lines  parallel  to  the  two  components;  then  AC  and  CB  represent  the  values 
(200  and  320  pounds  respectively)  and  the  directions  of  the  components. 

An  important  case  of  resolution  is  that  in  which  the  components  are  at 
right  angles  to  each  other.  Each  is  called  a  rectangular  component  or  re- 
solved part  of  the  force.  Rectangular  components  can  generally  be  com- 
puted more  easily  than  by  geometrical  construction.  Let  F  (Fig.  12)  be 
the  given  force  to  be  resolved  into  horizontal  and  vertical  components,  the 


9^ 

s 

F 

• 

/hk. 

b 

C=i^ 

A 

Fig.  10 

B 

c!a 

I 

> 

i    jJ^- 

r          1 

-i--~. 

b 

c^ 

Fig.  II 


angles  between  F  and  the  components  being  a  and  /5  respectively.  From 
ABC,  a  sketch  of  the  triangle  of  resolution,  not  necessarily  a  scale  drawing, 
it  is  plain  that  the  desired  components  equal  F  cos  a  and  F  cos  /3  respectively. 
And  always 


the  rectangular  com- 
ponent of  a  force 
along  any  line 


Jthe  magnitude  ^ 
■  tof  the  force       J  ^  " 


the  cosine  of  the 
acute  angle  between 
Ithe  force  and  that  line. 


The  components  of  a  force  along  two  rectangular  coordinate  axes  x  and  y 
are  called  the  x  and  y  components  of  the  force  respectively;  they  will  be 
denoted  by  Fj,  and  Fy. 

4.   Composition  of  Concurrent  Forces 

In  the  preceding  article  we  showed  how  to  determine  the  resultant  of  two 
concurrent  forces;  in  this  article  we  show  how  to  determine  the  resultant 
of  any  number  of  such  forces. 

§  I.  CoPLANAR  Forces. — Graphical  method.  By  means  of  the  paral- 
lelogram or  triangle  of  forces  (Art.  3)  find  the  resultant  R'  of  any  two  of 


12 


Chap,  i 


the  forces  of  the  given  set;  then  find  the  resultant  R"  of  any  other  given 
force  and  R';  then  the  resultant  of  another  given  force  and  R"\  and  so  on 
until  the  resultant  of  all  is  found.  Thus,  suppose  that  the  resultant  of  Fi, 
F2,  Fs,  and  Fi  (Fig.  13)  is  required:  Taking  the  given  forces  in  the  order 
in  which  they  are  numbered,  say,  we  first  draw  AB  parallel  to  Fi  and  equal 
to  Fi  by  some  convenient  scale,  then  BC  in  the  direction  of  and  equal  to 
F2;  then  AC  gives  the  magnitude  and  direction  of  R',  the  line  of  action  of 
R'  passing  through  O  parallel  to  AC.     Next  we  draw  CD  in  the  direction  of 


Fig.  13 

F3  and  equal  to  F3;  then  AD  gives  the  magnitude  and  direction  of  R",  the 
line  of  action  of  R"  passing  through  O  parallel  to  AD.  Next  we  draw  DE 
in  the  direction  of  and  equal  to  Fi;  then  AE  gives  the  magnitude  and 
direction  of  R'",  the  line  action  of  R'"  passing  through  0  parallel  to  AE. 
Of  course  the  lines  AC,  AD,  R',  and  R"  are  not  really  essential  to  the  solu- 
tion; they  were  drawn  here  and  referred  to  only  for  explanatory  purposes. 
The  force  polygon  for  a  set  of  forces  is  the  figure  formed  by  drawing  in 
succession  and  continuously  lines  which  represent  the  magnitudes  and 
directions  of   those   forces.     A   force   polygon    is    not   necessarily  a   closed 


Fig.  14 

figure;  thus  ABCDE,  not  including  EA,  is  a  force  polygon  for  Fi,  F2,  F3,  and 
Fi.  Many  force  polygons  can  be  drawn  for  a  given  set  of  forces,  as  many 
as  there  are  orders  of  taking  the  forces;  if  there  are  n  forces  in  the  set,  then 
I  •  2  •  3  •  •  •  '11  different  force  polygons  can  be  drawn.  In  Fig.  14  ad- 
ditional polygons  ABCDE  are  shown  for  Fi,  F2,  F3,  and  Fi  of  Fig.  13;  the 
lines  AE  represent  the  magnitude  and  direction  of  R.  The  bare  con- 
struction for  determining  the  resultant  of  a  set  of  concurrent  forces  can  now 
be  stated  thus:  Draw  a  polygon  for  the  forces;  join  the  beginning  and  the 
end  of  the  polygon,  and  draw  a  line  through  the  point  of  concurrence  of 
the  given  forces  parallel  to  the  joining  line;  the  joining  line,  with  arrow- 
head pointing  from  the  beginning  to  end  of  the  force  polygon,  represents 


Aet.  4 


13 


the  magnitude  and  direction  of  the  resultant,  and  the  other  line  its  line  of 

action. 

Algebraic  Method.  —  Choose  a  pair  of  rectangular  axes  of  resolution,  which 
let  us  call  X  and  y  axes,  with  origin  at  the  point  of  concurrence  of  the  forces 
to  be  compounded;  then  resolve  each  force  into  its  x  and  y  components 
at  the  origin,  and  imagine  it  replaced  by  them;  the  resulting  system  consists 
of  forces  in  ihex  and  in  the  3' axes;  next  find  the  resultant  of  the  forces  act- 
ing in  the  x  axis,  and  the  resultant  of  those  acting  in  the  y  axis;  finally,  get 
the  resultant  of  these  two  rectangular  resultants;  this  is  the  resultant  sought. 
For  example,  let  it  be  required  to  determine  the  resultant  of  the  six  forces 
acting  upon  the  4  foot  board  shown  in  Fig.  15.  The 
computations  in  outline  are  scheduled  below.  The 
values  of  the  angles  which  the  several  forces  make 
with  the  horizontal  were  computed  from  dimensions  in 
the  figure;  the  sum  of  the  x  components  is  +  3.40, 
and  that  of  the  3' components  is— 7.22  pounds.  The 
signs  of  the  sums  indicate  respectively  that  the  x  com- 
ponent of  the  resultant  R  acts  toward  the  right  and 
the  y  component  downward;  hence  the  resultant  acts 
to  the  right  and  downward.  The  angle  which  R  makes  with  the  horizontal 
is  tan~^  (7.22  -T-  3,40  =  2.123)  —  64°  47'-  The  value  of  the  resultant  is 
R  =  \/34o^  +  7.22^  =  7.98  pounds. 


5  lbs. 


F 

a 

cos  a 

sin  a 

Fx 

Fy 

8 

4 

6 

12 

7 
S 

0 

45° 
63°  26' 

36°  52' 
90° 
14°  2' 

I. 

0.707 

0.447 

0.800 

0. 

0.970 

0. 

0.707 
0.894 
0.606 
I . 
0.242 

+8. 

+  2.83 
-2.68 
—9.60 
0.00 
+4.85 

0 

+  2.83 
+  5-36 

—  7.20 

—  7.00 

—  I. 21 

+3 -40 

—  7.22 

§  2.  NoNCOPLANAR  FORCES.  —  Before  showing  how 
to  find  the  resultant  of  any  number  of  such  forces,  we 
explain  (i)  how  to  find  the  resultant  of  three  rectangular 
concurrent  forces  (lines  of  action  at  right  angles  to  each 
other),  and  (2)  how  to  resolve  a  force  into  three  nonco- 
planar  rectangular  components. 

(i)  If  three  noncoplanar  concurrent  forces  acting  on  a 
rigid  body  be  represented  by  OA ,  OB,  and  OC,  then  their 
resultant  is  represented  by  the  diagonal  OD  of  the  par- 
allelopiped  OABC  (Fig.  16).  (This  is  called  the  parallel- 
opiped  law,  and  OABC  is  called  a  parallelopiped  fof  forces.)  For,  according 
to  the  parallelogram  law,  OC  represents  the  resultant  of  two  of  the  forces  OA 


Fig.  16 


14 


Chap,  i 


and  OB,  and  OD  represents  the  resultant  of  OC  and  the  third  force  OC,  and 
(hence,  also)  the  resultant  of  the  three  given  forces.  This  law  leads  to  a 
simple  algebraic  method  for  finding  the  resultant  when  the  three  forces  are 
rectangular  (at  right  angles  to  each  other).  Thus,  let  Fi,  F2,  and  F3  (Fig.  17) 
be  the  three  forces,  R  their  resultant,  and  di,  62,  and  9$  the  angles  between  R 
and  the  forces  respectively;  then 

R'  =  Fi^  +  F2'  +  FzS 

cos  01  =  Fi/R,     cos  02  =  F2/R,      cos  ^3  =  F3/R. 

For  the  resultant  of  Fi  and  F2  (represented  by  OC,  Fig.  17)  equals 
(Fi^  +  Fa^)!,  and  hence  R^  =  (Fi^  +  F2'')  +  Fs'' ;  also  the  triangles  ODA, 
ODB,  and  ODC  are  right-angled  Sit  A,  B,  and  C  respectively,  and  hence 
cos^i  =  OA/OD  =  Fi/R,  cos  ^2  =  OB/OD  =  F2/R,  etc. 

(2)  A  force  can  be  resolved  into  three  noncoplanar  concurrent  forces  by 
applying  the  parallelepiped  law  inversely.  Thus,  let  OD  (Fig.  18)  represent 
the   given   force   F;    first,  construct  any  parallel opiped   of   which   OD  is  a 


Fig.  17 


Fig.  18 


Fig.  19 


diagonal;  then  the  three  edges  intersecting  at  0  represent  forces  equivalent 
to  the  given  force  because  the  resultant  of  these  three  forces  is,  according 
to  the  parallelepiped  law,  represented  by  OD.  Inasmuch  as  many  paral- 
lelopipeds  can  be  constructed  on  OD  as  diagonal,  many  sets  of  three  forces 
equivalent  to  the  given  force  can  be  found. 

The  practical  case  is  resolution  into  components  along  three  definite 
rectangular  axes;  then  there  is  only  one  set  of  components.  The  com- 
ponents may  be  found  quite  simply  by  an  algebraic  method:  thus,  let  F 
(Fig.  19)  be  the  force  to  be  resolved,  a,  13,  and  7  the  angles  between  F  and 
the  axes,  and  Fx,  Fy,  and  Fg  the  x,  y,  and  z  components  respectively;  then, 
since  OX,  OY,  and  OZ  are  projections  of  OD  on  the  rectangular  axes, 

Fx  =  F  cos  a,      Fy  =  F  cos  jS,      Fz  =  F  cos  7. 

Sometimes  the  direction  of  the  force  F  to  be  resolved  is  given  by  means  of 
two  angles,  one  being  the  angle  between  F  and  one  of  the  desired  compo- 
nents, and  the  other  being  the  angle  which  the  projection  of  F  on  the  plane 
of  the  other  two  components  makes  with  one  of  those  two,  as  for  instance 
a  and  4>  (Fig.  19).    Then  F  may  be  resolved  best  in  this  way:  first,  resolve 


Art.  4 


15 


it  into  two  components  F  cos  a  (along  the  x  axis)  and  F  sin  a  (in  the  plane  of 
the  y  and  2  axes),  and  then  resolve  F  sin  a  into  components  along  the  y 
and  2  axes,  that  is,  F  sin  a  sin  0  and  F  sin  a  cos  c^. 

^ny  number  of  noncoplanar  concurrent  forces  can  be  compounded 
graphically  by  means  of  their  force  polygon,  but  this  method  is  not  practi- 
cable generally,  because  the  polygon  is  not  a  plane  one;  however,  it  could 
be  drawn  in  "  plan  and  elevation  "  so  as  to  furnish  the  resultant  sought. 
The  algebraic  method  is  preferable;  it  is  carried  out  as  follows:  First, 
select  three  rectangular  axes  of  resolution  (here  called  x,  y,  and  2),  with 
origin  at  the  point  of  concurrence  of  the  forces  to  be  compounded;  next 
resolve  each  force  into  its  x,  y,  and  2  components,  and  imagine  it  replaced 
by  them,  thus  arriving  at  a  set  consisting  of  forces  acting  in  the  axes;  then 
find  the  resultants  of  the  forces  in  the  x,  in  the  y,  and  in  the  2  axis;  finally, 
compound  these  three  resultants,  thus  finding  the  resultant  sought. 

For  example,  let  it  be  required  to  determine  the  resultant  of  the  four 
forces  acting  on  a  4  foot  cube  (Fig.  20).  The  forces  are  concurrent  at  0; 
the  10  and  the  15  pound  forces  act  through  quarter 
points  of  certain  edges  as  shown.  The  x,  y,  and  2 
components  of  the  18  and  40  pound  forces  are  ob- 
viously as  scheduled  adjoining.  Since  the  15  pound 
force  is  perpendicular  to  the  x  axis,  its  x  component 
equals  zero;  and  since  the  angle  which  that  force 
makes  with  the  2  axis  =  tan~^  f  =  36°  52',  its  y 
and  2  components  are  15  sin  36°  52'  =  9,  and  15 
cos  36°  52'  =  12  pounds  respectively  as  scheduled. 
The  components  of  the  10  poimd  force  were  de- 
termined as  follows:  Since  Ya  =  5  and  YO  =  4  feet,  the  angle  which  the 
ID  pound  force  makes  with  the  y  axis  is  tan~^  |  =  51°  20';  the  y  component 
of  the  force  equals  10  cos  51°  20'  =  6.25  as  scheduled,  and  the  other  rec- 
tangular component  (in  the  zx  plane)  equals  10  sin  51°  20'  =  7.81  pounds. 


Fig.  20 


F 

Fz 

Fv 

F, 

18 
40 

IS 
10 

18.00 

0.00 

0.00 

-4.69 

0.00 

0.00 

—  9.00 

-6.25 

0.00 

40.00 

—  1 2 . 00 

-6.25 

+  13-31 

-15-25 

+  21.75 

The  angle  which  this  component,  acting  in  Ob,  makes  with  the  z  axis  equals 
tan~^  f  =  36°  52';  hence  the  x  and  z  components  of  the  10  pound  force 
equal  respectively  7.81  sin  36°  52',  or  4,69,  and  7.81  cos  36°  52',  or  6.25 
pounds.     The  value  of  the  resultant  is 

R  =  V  13.312  -f  15.252  +  21.752  =  29.7  pounds. 


l6  Chap,  i 

The  signs  of  the  sums  of  the  x,  y,  and  z  components  show  that  the  result- 
ant R  acts  toward  the  right,  downwards  and  forward.  Its  angles  with  the 
X,  y,  and  z  axes  are  respectively:  cos"'  (13.31  -r-  29.7)  =  63°;  cos~^  (15-25  -j- 
29.7)  =  59°;  cos-i  (21.75  -^  29.7)  =  43°. 


5.   Moment  of  a  Force;  Couples* 

§  I.  The  Moment  or  Torque  of  a  force  with  respect  to  a  point  is  the 
product  of  the  magnitude  of  the  force  and  the  perpendicular  distance  be- 
tween its  line  of  action  and  the  point.  The  perpendicular  distance  is  called 
the  arm  of  the  force  with  respect  to  that  point,  and  the  point  is  called  an 
origin  or  center  oj  moments.  Experience  suggests  the  notion  that  the 
moment  of  a  force  with  respect  to  a  point  is  a  measure  of  the  tendency  of 
the  force  to  rotate  the  body  about  a  line  through  the  point  and  perpen- 
dicular to  the  plane  of  the  force  and  the  point.  Such  a  notion  can  be 
verified  quite  accurately  by  means  of  a  simple  apparatus  represented  in 
Fig.  21.      It  consists  of  a  board  mounted  on  a  horizontal  shaft,  a  heavy  body, 

and  the  pail  which  can  be  suspended  horn  the 
board;  the  shaft  rests  in  ball  bearings  so  that 
practically  no  resistance  to  turning  is  exerted 
at  the  shaft;  the  board,  without  the  body  and 
the  pail,  is  well  balanced  so  that  gravity  would 
not  cause  it  to  turn  from  any  position.  Now, 
let  the  pail  containing  shot  be  hung  from  B,  C,  D, 
etc.,  in  succession,  the  amount  of  shot  being 
taken  so  that  the  heavy  body  will  be  supported,  OA  not  being  horizontal 
necessarily.  Then  in  each  case  the  turning  effect  of  the  pull  at  B,  C,  or 
D  equals  the  turning  effect  of  the  pull  at  A ;  hence  the  turning  effects  of 
the  pulls  at  B,  C,  D,  etc.,  are  equal.  And  if  the  moments  of  these  pulls 
(several  weights  of  pail  and  shot)  about  O  be  computed,  then  those  mo- 
ments will  be  found  equal  too,  and  therefore  moments  are  measures  of  turn- 
ing effects. 

It  follows  from  the  definition  of  moment  that  the  unit  moment  is  that  of 
a  unit  force  whose  arm  is  a  unit  length.  There  are  no  one-word  names  for 
any  of  these  units  of  moment;  the  units  are  called  foot-pound,  inch- ton,  etc., 
according  as  the  unit  length  and  force  are  the  foot  and  the  pound,  the  inch 
and  the  ton,  etc. 

In  a  discussion  involving  the  moments  of  several  forces,  it  is  generally 
convenient  to  give  signs  to  the  moments  to  indicate  the  directions  (clock- 
wise or  anticlockwise)  in  which  the  several  forces  turn  or  tend  to  turn 
the  body  to  which  they  are  applied  about  the  origin  in  question.  In  this 
book,  clockwise  rotation  is  regarded  as  negative  and  anti  as  positive,  and 
rotations  are  supposed  to  be  viewed  from  the  reader's  side  of  the  printed  page; 

*  See  Art.  8  also. 


Art.  s 


17 


lOOlbs. 


30 


A 

JB... 

■--  — 
■s. 

thus  the  moment  of  the  200  pound  force  (Fig.  22)  about  O  is  positive  and 
about  A  negative. 

Principles  of  Moments.  —  If  two  sets  of  coplanar 
forces  are  equivalent  (Art.  2),  then  the  moment-sum* 
for  one  set  with  respect  to  any  point  equals  that  for 

the  other  with  respect  to  the  same  point.     This  will  \     A       \         ^ 

be  granted  as  self-evident  by  most  students;  others 
may  be  convinced  by  this:  Let  Si  and  ^'2  denote  the 
two  equivalent  sets,  and  Sz  a  third  set  (coplanar  with  ''200 lb5. 

Si  and  ^2)  which  could  balance  Si,  and  hence  also  S2.  yig.  22 

Since  Si  and  Sz  would  balance,  they  together  would  not 

turn  the  body  on  which  they  act  about  any  line;  hence  the  moment-sums  for 
Si  and  53  with  respect  to  any  point  0  in  the  plane  are  equal  in  value  but 
opposite  in  sign.  Likewise,  the  moment-sums  for  S2.  and  ^3  with  respect  to  O 
are  equal  in  value  but  opposite  in  sign.  Therefore  the  moment-sums  for  Si  and 
S2  wath  respect  to  O,  being  equal  to  the  same  thing,  are  equal.  It  follows  from 
the  foregoing  principle  that  the  moment-sum  for  any  set  of  coplanar  forces 
with  respect  to  any  point  in  their  plane  equals  the  moment  of  their  resultant 
with  respect  to  that  point.  Also,  the  moment  of  a  force  with  respect  to  a 
point  equals  the  moment-sum  of  its  components  with  respect  to  the  same 
point,  the  components  to  be  coplanar  with  the  given  force  and  the  point. 

When  the  moment  of  a  force  about  a  certain  point  must  be  computed, 
and  the  arm  of  the  force  with  respect  to  that  point  cannot  be  easily  de- 
termined, then  the  desired  moment  can  be  computed,  more  easily  perhaps, 
from  components  of  the  force,  by  aid  of  the  preceding  principle.  Thus, 
let  it  be  required  to  compute  the  moment  of  the  100  pound  force  (Fig. 
22)  about  O.  The  horizontal  and  vertical  components  of  the  force  are 
respectively,  86.6  and  50  pounds;  imagining  them  applied  at  A  makes 
their  arms  3  and  4  feet  respectively,  hence  the  desired  moment  equals 
—  (86.6  X  3)  —  (50  X  4)  =  —  459-8  foot-pounds.  Sometimes  the  components 
can  be  applied  (in  imagination)  so  that  one  passes  through  the  origin  of 
moments;  then  its  moment  equals  zero  and  the  desired  moment  equals  the 
moment  of  the  other  component.  Thus  the  horizontal  and  vertical  compo- 
nents of  the  200  pound  force  equal  89.4  and  178.8  pounds  respectively; 
imagining  them  applied  at  C,  their  arms  are  o  and  3  feet,  and  so  the  desired 
moment  equals  178.8  X  3  =  536.4  foot-pounds. f 

*"  Moment-sum  "  means  the  algebraic  sum  of  the  moments  of  the  forces  of  a  system; 
torque  of  a  set  of  forces  means  the  same  thing. 

t  Some  writers  regard  the  parallelogram  law  (for  forces)  as  fundamental,  and  deduce 
the  principle  of  moments  from  the  law.  We  give  such  a  deduction  of  the  principle  for  two 
forces,  —  namely,  the  moment  of  the  resultant  of  two  concurrent  forces  about  any  point  in 
their  plane  equals  the  algebraic  sum  of  the  moments  of  the  forces  about  the  same  point 
(Varignon's  theorem).  The  theorem  can  be  extended  easily  to  any  number  of  coplanar 
forces,  thus  proving  the  principle. 

Let  P  and  Q  (Fig.  24)  be  two  concurrent  forces  acting  —  on  a  body  not  shown  —  in 


i8 


Chap, 


b 

B 

r"^ 

c 

% 

V 

a 

A 

V 

Fig.  23 


§  2.  Couples  (see  also  Art.  8),  —  Two  equal  and  parallel  forces  which 
are  opposite  in  sense  may  advantageously  be  considered  together;  so  con- 
sidered, they  are  called  a  couple.  By  arm  of  the  couple  is  meant  the  per- 
pendicular distance  between  the  lines  of  action  of  the  forces;  by  plane  of 
the  couple  is  meant  the  plane  of  those  lines  of  action.  The 
jnoment  or  torque  of  the  couple  about  any  point  in  the  plane  of 
the  couple  is  the  algebraic  sum  of  the  moments  of  the  forces 
about  that  same  point.  This  sum,  or  moment,  including  the 
sign,  is  the  same  for  all  origins  in  the  plane.  For,  let  F  and  F 
(Fig.  23)  be  the  forces  of  the  couple,  AB  being  the  arm;  then 
the  moments  of  the  couple  with  respect  to  the  origins  a,  b,  and 
c  are  respectively:  -  F(Aa)  +  FiBa)  =  F{AB);  +  F{Ab)  - 
F(Bb)  =  F(AB);  and  +  F{Ac)  +  F(Bc)  =  F(AB).  Since  a,  b, 
and  c  represent  all  possible  origins,  the  proposition  is  proved. 
Since  the  origin  of  moments  is  immaterial,  no  reference  is  made  to  it  in 
speaking  of  the  moment  of  a  couple.  Moreover,  the  value  of  the  moment  is 
computed  more  simply  than  as  stated  in  the  definition,  by  multiplying  the 
common  magnitude  of  the  forces  and  the  arm  of  the  couple.  As  just  shown, 
the  moment  of  the  couple  FF  is  positive ;  that  is,  the  couple  would  turn  the 
body,  if  free,  counterclockwise  about  any  axis  perpendicular  to  the  plane  of 
the  couple;  obviously,  the  moment  of  the  couple  GG  is  negative.  Sense  of  a 
couple  refers  to  the  way  in  which  the  couple  would  turn  a  body;  thus  we 
speak  of  a  clockwise  sense  or  a  counterclockwise  sense. 

Two  coplaner  couples  whose  mojnents  (including  sign)  are  equal  are 
equivalent.  The  following  proof  is  in  two  parts;  namely,  (i)  when  the 
four  forces  of  the  couples  are  nonparallel,  and  (2)  when  they  are  paral- 
lel, (i)  Let  P1P2  (Fig.  25)  be  one  given  couple,  and  Q1Q2  the  other,  their 
arms  being  p  and  q  respectively;  then  Pp  =  Qq.  We  will  show  that  the 
P  couple  would  balance  the  reversed  Q  couple;  it  will  follow  that  the 
given  couples  are  equivalent.  The  area  of  the  parallelogram  A  BCD  = 
ADp  =  ACq;  therefore,  since  Pp  =  Qq,  AD/ AC  =  P/Q.  That  is,  the  sides 
of  the  parallelogram  represent  the  forces  P  and  Q;   and  so  the  diagonal 


the  lines  OA  and  OB  respectively,  and  let  22  be  their 
resultant.  Also,  let  OABC  be  a  parallelogram  for  the 
forces,  D  any  origin  of  moments,  and  a,  13,  and  7  the 
angles  between  OD  and  the  forces  respectively  as 
marked.  Now  P  sin  a,  Q  sin  /3,  and  R  sin  7  are  the 
values  of  the  components  of  P,  Q,  and  R  at  right  angles 
to  OD,  and  it  is  clear  from  the  figure  that  P  sin  a  -\- Q 
sin  0  =  R  sin  7.  Multiplying  through  by  OD,  we 
get  P  •  OD  sin  a  +  Q  •  OD  sin  0  =  R-OD  sin  7.  But 
these  terms  are  the  moments  of  P,  Q,  and  R  respec- 
tively; hence  the  moment  of  R  equals  the  sum  of  the 

moments  of  P  and  Q  as  stated.     When  D  is  between  P  and  Q,  then  a  slight  modification  in 
the  proof  is  necessary. 


Art.  s 


19 


AB  represents  the  resultant  of  Qi  reversed  and  Pi,  and  the  diagonal 
BA  represents  the  resultant  of  Q2  reversed  and  P2.  Since  the  resultants  are 
equal,  opposite,  and  colinear  they  balance,  and  so  the  P  couple  and  the 
reversed  Q  couple  balance.  Hence,  etc.  (2)  When  Pi,  P2,  Qi  and  Q2  are 
parallel,  and  the  moments  of  the  two  couples  are  equal,  then  each  couple  is 
equivalent  to  some  third  couple,  the  forces  of  which  intersect  Pi,  P2,  Qi, 
and  Qi,  according  to  (i).     Therefore  they  are  equivalent  to  each  other. 

§  3.  A  Force  and  a  Couple.  —  The  resultant  of  a  coplaner  force  and  couple 
is  a  single  force;  the  resultant  is  equal  to  and  has  the  same  direction  as  the  force, 
and  its  moment  about  any  point  on  the  given  force  equals  the  moment  of  the  couple. 
Proof  follows: 

Let  F  (Fig.  26)  be  the  given  force,  and  P1P2  the  given  couple.  (If  the 
forces  of  the  given  couple  are  parallel  to  F,  then  imagine  the  couple  shifted 


Fig.  25 


Fig.  26 


until  they  are  not  so  parallel.)  Now  suppose  that  AB  and  BC  represent 
the  magnitudes  and  directions  of  Pi  and  F  respectively;  then  AC  repre- 
sents the  magnitude  and  direction  of  the  resultant  of  those  two  forces. 
(The  line  of  action  of  the  resultant  is  R' ,  parallel  to  ylC  and  through  the  in- 
tersection of  Pi  and  F.)  Let  CD  equal  AB;  then  AD  represents  the  magni- 
tude and  direction  of  the  resultant  of  R'  and  P2,  and  hence  of  the  three 
forces  Pi,  F,  and  P2.  But  AD  is  equal  and  parallel  to  BC;  hence  this  final 
resultant  is  equal  and  parallel  to  F.  (The  line  of  action  of  this  final  resultant 
is  R,  parallel  to  BC  and  through  the  intersection  of  R'  and  P2.)  Since  R  is 
equivalent  to  F,  Pi,  and  P2,  its  moment  about  any  point  of  F  equals  the  sum 
of  the  moments  of  F,  Pi,  and  P2  about  that  point;  but  F  has  no  moment 
about  such  point,  and  hence  the  moment  of  R  equals  the  sum  of  the  moments 
of  Pi  and  P2  (the  moment  of  the  couple). 

It  follows  from  the  foregoing  that  a  force  R  can  he  resolved  into  a  force  equal 
and  parallel  to  R,  and  a  couple  whose  moment  equals  that  of  R  about  any  point 
on  the  component  force.  Thus  the  moment  of  the  couple  component  depends 
on  the  line  of  action  chosen  for  the  force  component.  Independent  proof 
of  this  proposition  follows : 


20 


Chap,  i 


Fig. 


27 


Let  R  (Fig.  27)  be  the  force  to  be  resolved,  and  O  a  point  through  which 
the  line  of  action  of  the  force  component  is  to  pass.     First  we  resolve  R 

into  two  concurrent  components,  one  of  which 
passes  through  0;  take  any  point  on  R  (as  a) 
for  the  point  of  concurrence  and  any  direction 
(as  ab)  for  the  line  of  action  of  the  second  com- 
ponent. These  components  we  call  Ci  and  C2 
respectively.  To  determine  Ci  and  C2,  we  draw 
AB  to  represent  R,  and  AC  and  BC  parallel 
to  Ci  and  Co  respectively;  then  AC  =  Ci,  and 
CB  =  C2.  Next  we  resolve  Ci  at  O  into  two 
components  parallel  to  C2  and  R,  which  com- 
ponents we  call  C3  and  C4  respectively.  To 
determine  Cs  and  d,  we  draw  from  A  a  line 
parallel  to  C3  and  from  C  a  line  parallel  to  d, 
and  so  locate  D;  then  AD  =  C3,  and  DC  =  C4.  Ob\'iously  now  C2,  C3  and 
d  are  equivalent  to  R,  that  is,  they  are  components  of  R;  and  as  required 
d  passes  through  0,  and  C2  and  d  (equal,  parallel,  and  opposite)  constitute 
a  couple.  Moreover,  according  to  the  principle  of  moments,  the  moment  of 
R  about  any  point  on  d  equals  that  of 
C2,  C3,  and  d  about  that  point;  but  the  '"'"''^^^f^^M^ 
moment  of  d  equals  zero,  hence,  etc.* 

6.   Graphical  Composition  of   Coplanar 
Nonconcurrent  Forces 

§  I.  First  Method.  —  When  the  forces 
to  be  compounded  are  not  parallel  nor 
nearly  so,  then  we  compoimd  any  two  of 
the  forces,  next  their  resultant  and  the 
third  force,  that  resultant  and  the  fourth 
force,  and  so  on  until  the  resultant  of  all 
the  forces  has  been  found.  For  example, 
consider  the  forces  acting  on  the  retaining  wall  shown  in  section  in  Fig.  28: 

*  (i)  Composition  of  a  Force  and  a  Couple  and  (2)  Resolution  of  a  Force  into  a  Force  and 
a  Couple  can  be  performed  also  as  follows  (student  should  supply  figure):  (i)  Replace  the 
couple  by  an  equivalent  couple  whose  forces  equal  the  given  force,  and  place  the  couple  so 
that  one  of  its  forces  is  colinear  with  and  opposite  to  the  given  force.  These  two  forces 
balance;  the  other  force  of  the  new  couple  remains,  and  it  is  the  resultant  sought.  (Study 
of  the  steps  in  the  process  shows  that  the  resultant  force  is  equal  and  parallel  to  the  original 
force,  and  that  the  moment  of  the  resultant  about  a  point  on  the  line  of  action  of  the  original 
force  equals  the  moment  of  the  couple.)  (2)  Apply  two  forces  at  the  given  point  equal  and 
parallel  to  the  given  force  and  opposite  to  each  other.  These  two  forces  along  with  the 
given  force  can  be  grouped  into  a  force  and  a  couple,  and  they  (the  force  and  couple)  are 
the  components  sought.  (Study  of  the. steps  of  the  process  shows  that  the  component 
force  is  equal  and  parallel  to  the  given  force,  and  the  moment  of  the  couple  equals  that  of 
the  given  force  about  the  given  point.) 


Fig.  28 


Art.  6 


21 


they  consist  of  its  own  weight  (16,000  pounds  per  foot  of  length),  the  earth 
pressure  on  the  back  (6000  pounds),  that  on  the  top  of  the  base  (9000  pounds), 
and  that  on  the  bottom  of  the  base.  The  resultant  of  the  first  three  forces 
will  now  be  determined.  We  draw  AB  and  BC  to  represent  the  6000  and 
the  16,000  pound  forces,  and  then  join  A  and  C;  AC  represents  the  magni- 
tude and  direction  of  the  resultant  of  the  two  forces,  and  the  line  marked 
R'  (parallel  to  AC  and  through  point  i)  is  the  line  of  action  of  that  resultant. 
We  next  draw  CD  to  represent  the  9000  pound  force,  and  join  A  and  D; 
AD  represents  the  magnitude  and  direction  of  the  resultant  of  R'  and  9000 
(and  hence  also  of  the  three  given  forces),  and  the  line  marked  R  (through 
point  2  and  parallel  to  AD)  is  the  line  of  action  of  that  resultant. 

It  may  be  noted  that  the  magnitude  and  the  direction  of  the  resultant  is 
found  just  as  for  concurrent  forces  (Art.  4).  For  nonconcurrent  forces  it 
is  necessary  to  draw  the  lines  of  action  of  the  intermediate  resultants 
(R',  R",  etc.),  in  order  to  find  the  line  of  action  of  the  final  resultant,  lines 
which  are  unnecessary  when  compounding  concurrent  forces. 

When  the  forces  are  parallel  or  nearly  so,  the  foregoing  method  fails 
because  there  is  no  accessible  intersection  of  the  lines  of  action  of  two 
given  forces  through  which  to  draw  the  line  of  action  of  the  first  resultant. 
This  difiiculty  can  be  met  as  follows:  Introduce  into  the  given  system 
two  equal,  opposite,  and  colinear  forces,  which  will  not  change  the  resultant, 
taking  their  common  line  of  action  somewhat  across  those  of  the  given 
forces;  then  use  the  first  method,  compounding  first  any  pair  of  forces 
whose  intersection  is  accessible,  etc. 

§  2.  Second  Method,  applicable  to  any  coplanar  forces.  —  We  first  re- 
solve each  force  into  two  concurrent  components,  resolving  in  such  a  way 


Fig.  29 

that  these  components,  excepting  one  of  the  first  force  and  one  of  the  last 
force,  balance  or  destroy  each  other;  these  two  remaining  components 
are,  in  general,  concurrent,  and  so  we  readily  find  their  resultant,  which  is 
also  the  resultant  of  the  given  forces.  For  example,  let  Fi,  F2,  F3,  and  F4 
(Fig.  29)  be  the  forces  to  be  compounded.  First  we  draw  a  force  polygon 
for  the  given  forces,  taking  them  in  any  convenient  order,  as  ABCDE; 
then  we  take  any  convenient  point  O  as  the  common  vertex  of  the  tri- 
angles of  resolution.  AO  and  OB  represent  two  components  of  Fi  in  mag- 
nitude and  direction,  BO  and  OC  two  components  of  F2,  etc.;    thus  this 


22  Chap,  i 

resolution  gives  several  pairs  of  equal  and  opposite  components,  OB  and 
BO,  OC  and  CO,  OD  and  DO.  The  components  of  Fi  are  taken  to  act 
through  point  i,  those  of  F^  through  2,  those  of  Fz  through  3,  etc.,  ti 
first  point,  i,  being  taken  at  pleasure  on  Fi,  point  2  where  oh  intersects  F2, 
point  3  where  oc  intersects  7^3;  etc.  Thus  the  components  OB  and  BO  are 
colinear  and  they  balance;  likewise  OC  and  CO,  and  OD  and  DO.  Only  the 
first  and  last  components  AO  and  OE  remain;  their  resultant  is  represented 
hy  AE  in  magnitude  and  direction,  and  its  line  of  action  is  ae  (parallel  to 
AE  through  the  intersection  of  ao  and  oe). 

The  common  vertex  of  the  triangles  of  resolution  0  (Fig.  29)  is  the  pole 
of  the  force  polygon;  the  lines  from  the  pole  to  the  vertexes  of  the  force 
polygon,  0/1,  OB,  OC,  etc.,  are  rays;  the  line  of  action  of  the  several  forces, 
oa,  ob,  OC,  etc.,  are  strings  which,  considered  collectively,  is  the  string  or 
funicular  polygon  (also  called  equilibrium  polygon,  especially  when  the  given 
forces  are  balanced  or  in  equilibrium).  The  rays  are  sometimes  referred 
to  by  number,  OA  being  the  first,  05  the  second,  etc.;  likewise  the  strings. 

In  using  this  second  method,  the  beginner  had  best  reason  out  the  vari- 
ous steps  of  the  construction  somewhat  as  in  the  foregoing.  After  some 
practice  he  might  use  the  following  aids:  (i)  The  two  strings  intersecting 
on  the  line  of  action  of  any  force  are  parallel  to  the  rays  drawn  to  the  ends 
of  that  side  of  the  force  polygon  corresponding  to  that  force,  thus  the  strings 
intersecting  on  be  are  ob  and  oc.  (2)  The  string  which  joins  points  in  the 
lines  of  action  of  any  two  forces  is  parallel  to  the  ray  which  is  drawn  to 
the  common  point  of  the  two  sides  of  the  force  polygon  corresponding  to 
those  forces,  or,  the  string  joining  points  on  be  and  cd  is  parallel  to  OC. 
(3)  The  bare  construction  in  the  second  method  is  simply  this:  Draw  a 
force  and  a  string  polygon  for  the  forces,  then  draw  a  line  from  the  begin- 
ning to  the  end  of  the  force  polygon  and  a  parallel  line  through  the  inter- 
section of  the  first  and  last  strings;  the  first  line  represents  the  magnitude 
and  direction  of  the  resultant  (sense  being  from  the  beginning  to  the  end  of 
the  force  polygon),  and  the  second  line  is  the  line  of  action  of  the  resultant. 

This  second  method  is  not  so  simple  in  principle  as  the  first,  but  in  the 
second  there  is  more  opportunity  for  varying  the  construction  to  keep 
the  drawing  within  convenient  limits;  thus  the  pole  may  be  shifted,  and  the 
starting  point  of  the  string  polygon  may  be  taken  anywhere  on  any  of  the 
given  forces.  Though  many  string  polygons  may  be  drawn  for  a  given  set 
of  forces,  all  determine  the  same  line  of  action  of  the  resultant;  that  is,  the 
intersections  of  the  first  and  last  strings  of  all  string  polygons  lie  on  one 
straight  line,  the  line  of  action  of  the  resultant. 

§  3.  When  the  Force  Polygon  Closes. — It  may  seem,  at  first  thought, 
that  the  resultant  vanishes,  or  is  zero;  in  general,  this  conclusion  would 
be  wrong,  the  system  actually  reducing  to  a  couple.  Thus,  let  Fi,  F2,  F3, 
and  Fi  (Fig.  30)  be  a  force-set  whose  force  polygon  ABODE  closes;  using 
the  first  method  for  compounding,  we  find  that  the  resultant   R"  of  the 


Art.  7 


23 


first  three  forces  is  given  by  ^D  in  magnitude  and  in  direction,  and  ad 
is  its  l^e  of  action;  hence* i?"  is  equal,  opposite,  and  parallel  to  F4,  and  so 
t^  ,  given  force-set  reduces  to  a  couple  (R",  Fi).  The  arm  of  this  couple  is 
tne  perpendicular  distance  between  Ft  and  R",  and  so  the  moment  of  the 
couple  is  the  product  of  Fi  (or  R")  and  the  arm  (according  to  the  scale 
of  the  space  diagram) ;  the  sense  of  the  couple,  clockwise,  is  apparent  from 
the  relative  positions  and  directions  of  the  forces  of  the  couple  as  seen  in 
the  space  diagram.  In  Fig.  31  the  composition  has  been  made  by  the 
second  method;  the  system  reduces  to  the  two  components  AO  (acting  in 
ao)  and  OE  (acting  in  oe).  These  components  are  equal,  opposite,  and 
parallel,  and  so  the  given  force-set  reduces  to  a  couple.  The  arm  of  the 
couple  is  the  perpendicular  distance  between  the  first  and  last  strings,  ao 
and  oe;  the  moment  of  the  couple  is  the  product  of  OA  or  EO  (according 


E      ^= .— ^D  i"^^ J^A^ 

Fig.  30  Fig.  31 

to  the  scale  of  the  force  diagram)  and  the  arm  (according  to  the  scale  of 
the  space  diagram) ;   the  sense  is  apparent  from  the  space  diagram. 

The  length  of  the  arm  and  the  magnitude  of  the  forces  of  the  couple 
depend  on  the  order  in  which  the  forces  are  taken  in  the  force  polygon,  in 
the  first  method;  and  upon  the  position  chosen  for  the  pole  0,  in  the  second 
method.  But  the  moment  of  the  couple  is  independent  of  all  these  vari- 
ations. This  fact  may  be  verified  by  actually  compounding  a  certain  force- 
set  (whose  force  polygon  closes)  in  several  ways,  making  all  these  different 
variations  and  thus  arriving  at  different  couples.  The  couples  are  all 
equivalent  to  the  same  force-set  and  so  equivalent  to  each  other,  and 
hence  their  moments  are  equal  (Art.  5). 

7.  Algebraic  Composition  of  Coplanar  Nonconcurrent  Forces 

§  I.  Parallel  Forces.— If  the  forces  be  given  sign,  those  in  either  direc- 
tion being  called  positive  and  those  in  the  other  negative,  then  the  alge- 
braic sum  of  the  forces  gives  the  magnitude  and  sense  of  the  resultant,  the 
sign  of  the  sum  indicating  the  sense  of  the  resultant.  According  to  the 
principle  of  moments  (Art.  5),  the  moment  of  the  resultant  about  any  point 
equals  the  algebraic  sum  of  the  moments  of  the  forces  about  that  point,  and 


24 


Chap,  i 


20  lbs. 

^30  It 

> 

'40  lbs.                    1 50  lbs. 

Fig.  32 


this  requirement  fixes  the  position  or  line  of  action  of  the  resultant.  For 
example,  let  us  find  the  resultant  of  the  four  forces  acting  on  a  10  foot 
board,  as  shown  in  Fig.  32.  CalUng  upward  forces  positive,  the  alge- 
braic sum  is  +  20  —  40  —  50  + 30  =—40;  hence  the  resultant  equals  40 
pounds  and  acts  downward.     The  algebraic  sum  of   the  moments  of  the 

forces  about  the  left  end  of  the 
^^30 lbs.  board,  say,  is  o  —  120  —  350  +  270  = 
—  200  foot-pounds,  and  hence  the 
moment  of  the  resultant  also  is  —  200 
foot-pounds;  this  fixes  the  arm  of  the 
resultant,  200  H-  40,  or  5  feet.  Since 
the  resultant  acts  downwards  and  its  moment  about  the  origin  is  negative, 
its  line  of  action  must  be  to  the  right  of  the  origin,  5  feet. 

To  find  the  resultant  of  two  parallel  forces  (a  common  problem),  we 
may  of  course  use  the  general  method  just  explained,  but  the  following 
special  results  are  worth  noting.  We  distinguish  two  cases:  (i)  the  two 
forces  are  alike  in  sense;  (2)  they  are  opposite.  In  (i)  the  resultant  equals 
the  sum  of  the  forces  and  agrees  with  them  in  sense;  in  (2)  the  resultant 
equals  the  difference  between  the  two 
forces  and  agrees  with  the  larger  in 
sense.  In  order  that  the  moment  of 
the  resultant  R  may  equal  the  sum  of 
the  moments  of  the  forces,  P  and  Q 
(Fig.  33),  then,  in  case  (i),  R  must  lie 
between  the  forces,  and  in  case  (2)  out- 
side of  them  and  adjacent  to  the  larger  force  (assumed  to  be  P  in  the  figure). 
Furthermore,  if  the  distances  from  R  to  P  and  Q  be  called  p  and  g  respec- 
tively, and  that  between  P  and  Q  be  a,  then  in  either  case,  Rp  =  Qa  and 
Rg  =  Pa,  or  p  =  Qa/R     and     g  =  Pa/R 

either  of  which  definitely  fixes  the  position  of  R.  Also  for  either  case,  Pp 
=  Qg    or    P/Q  =  g/p;  hence 

P/Q  =  BC/AC, 

that  is,  the  line  of  action  of  the  resultant  of  two  parallel  forces  divides  any 
secant  intersecting  their  lines  of  action  into  two  segments  which  are  inversely 
proportional  to  the  two  forces. 

If  the  algebraic  sum  of  a  set  of  parallel  forces  equals  zero,  then  it  may 
appear  to  the  student  that  their  resultant  vanishes  or  is  zero;  this  does 
not  follow,  but  the  resultant  actually  is  in  general  a  couple.  For  the  re- 
sultant of  all  but  one  of  the  given  forces  is  a  single  force  equal,  opposite, 
and  parallel  to  the  omitted  one;  but  these  two  are  not  in  general  colinear, 
and  so  they  constitute  a  couple,  the  resultant  of  the  system.  The  couple 
arrived  at  depends  on  which  one  of  the  given  forces  is  omitted,  but  the 
moment  of  the  couple  does  not,  for  that  couple  is  the  resultant  of  the  set, 


hP-H 

r;  -q->l 

C 

> 

<- 

q -> 

^i 

B 

<    --ip  ■) 

A    i 

P 

R 

^R           Ip 

a 

Fig.  33 


Art.  7 


25 


A60lb&. 


401bvf 


"20  lbs. 


SOlbs. 
Fig.  34 


SOlbs. 


and  the  moment  equals  the  algebraic  sum  of  the  moments  of  given  forces, 
a  definite  quantity.  For  example,  let  us  find  the  resultant  of  the  five 
forces  acting  on  a  10  foot  board,  as  shown  in  Fig.  34.  Their  algebraic  sum  is 
zero,  and  so  their  resultant  is,  presumably,  a  couple.  Compounding  all  but 
the  40  pound  force,  we  find  that  their 
resultant  equals  40  pounds,  acts  down- 
ward, 7.5  feet  to  the  right  of  the  left 
end  of  the  board,  and  so  the  resultant 
is  a  couple  whose  moment  is  +  (40  X 
2.5)  =  -|-ioo  foot-pounds. 

Instead  of  actually  determining  the  forces  of  the  resultant  couple  as 
explained,  it  is  usually  sufficient  to  determine  the  moment  of  the  result- 
ant couple;  this  moment  equals  the  algebraic  sum  of  the  moments  of  the 
given  forces  about  any  point.  Thus,  in  the  preceding  example,  after  ascer- 
taining that  the  resultant  is  a  couple,  we  compute  the  moment-sum  for  the 
given  forces,  wdth  moment  origin  at  the  middle  of  the  board,  say,  or  (20  X  5) 
-  (60  X  3)  +  (30  X  i)  -  (50  X  i)  +  (40  X  5)  =  +100  foot-pounds;  and 
then  conclude  that  any  couple  whose  moment  equals  -fioo  foot-pounds 
may  be  regarded  as  the  resultant  of  the  system. 

§  2.  NoNPARALLEL  FORCES.— As  shown  in  Art.  6,  the  resultant  is  in  general 
a  single  force,  given  in  magnitude  and  direction  by  the  line  joining  the  be- 
ginning and  end  of  the  force  polygon  for  the  forces.  It  follows,  therefore, 
that  the  component  of  that  resultant  force  along  any  line  equals  the  alge- 
braic sum  of  the  components  of  the  given  forces  along  that  line.  From  this 
principle  we  can  get  the  components  of  the  resultant  along  any  two  rec- 
tangular axes;  and  from  these  components  the  magnitude  and  direction  of 
the  resultant  itself  can  be  readily  determined  by  obvious  means.  Ac- 
cording to  the  principle  of  moments  (Art.  5),  the  moment  of  the  resultant 
about  any  point  must  equal  the  sum  of  the  moments  of  the  given  forces  about 
that  point;  and  this  requirement  fixes  the  position  or  line  of  action  of  the 
resultant.     For  example,  let  us  find  the  resultant  of  the  six  forces  acting 

on  a  board,  4  by  4  feet,  as  shown  in  Fig.  35.  The 
angles  which  the  forces  make  with  the  horizontal  and 
the  arms  of  the  forces  with  respect  to  the  center  of 
the  board  are  recorded  in  columns  2  and  3  of  the 
schedule  on  page  26;  they  could  be  computed  trigono- 
metrically  or  could  be  scaled  from  a  larger  drawing. 
The  X  and  y  components  of  the  several  forces  are 
recorded  in  columns  4  and  5  respectively,  and  the 
moments  of  the  forces  with  respect  to  the  center  of 
the  board  in  column  6.     The  algebraic  sums  of  the  x 


^6lbs. 


VIbs. 


51b5^ 


Fig.  35 

y  components  are  +3.40  and   -7.22  pounds  respectively;   hence 


and  the 

R  =  V'3.40^  -|-  7.22^  =  7.98  pounds 


The  signs  of  the  sums  indicate  that  R 


acts  toward  the  right  and  downward;    the  angle  which  R  makes  with  the 


26 


Chap. 


horizontal  is  tan~^  (7.22  -J-  3.40),  or  64°  47'.  The  sum  of  the  moments  is 
—  14.14  foot-pounds;  and,  since  the  moment  of  R  also  equals  —14.14,  R  lies 
on  the  right-hand  side  of  the  origin  of  moments  (the  moment  being  negative), 
and  its  arm  is  14.14  -^  7.98  =  1.77  feet.  Thus,  R  has  been  completely 
determined. 


I 

2 

3 

4 

s 

6 

7 

8 

9 

10 

F 

a 

a 

F. 

Fy 

M 

fli 

02 

Ml 

il/2 

8 

0 

I  .00 

-1-8.00 

0.00 

-    8.00 

I 

-    8.00 

0 

4 

45° 

0.  71 

+  2.83 

+  2.83 

-    2.84 

I 

0 

-    2.83 

0 

6 

63°  25' 

0.89 

-2.68 

+  5.36 

-  S-37 

0 

I 

0 

-    5-36 

7 

90° 

2.00 

0.00 

—  7.00 

-I-14.00 

2 

0 

-t-14.00 

12 

36°  52' 

I  .60 

—  9.60 

—  7.20 

—  19.20 

2 

0 

—  19.20 

0 

5 

14°  2' 

1-45 

+4.85 

—  I. 21 

+   7-27 

I 

2 

+   4.85 

+    2.42 

+3 -40 

—  7.22 

-14.14 

-25.18 

-f  1 1 .  06 

The  moment-sum  (—14.14)  may  be  determined  also — more  simply  in 
this  example  —  by  adding  the  moments  of  the  x  and  y  components  of  the 
several  forces  (Art.  5).  This  method  requires  that  we  take  definite  lines 
of  action  for  all  the  components.  Of  course  any  force  and  its  x  and  y  com- 
ponents must  be  concurrent  (Art.  3).  We  take  the  heavy  dots  (Fig.  35), 
as  the  points  of  concurrence;  then  the  arms  of  the  x  components  of  the 
forces  respectively  with  respect  to  the  center  of  the  board  are  as  recorded 
in  column  7,  and  the  arms  of  the  y  components  are  as  recorded  in  col- 
umn 8.  The  moments  of  the  x  components  and  of  the  y  components  are 
recorded  in  columns  9  and  10  respectively.  The  sums  of  the  moments  of 
the  X  components  and  of  the  y  components  are  —25.18  and  +11.06  foot- 
pounds respectively,   and  hence   the  moment-sum  for   the   given   forces  is 

—  25.18+  11.06  =  —  14.12. 

If  2/^x  and  ^Fy  for  a  force-set  both  equal  zero,  then  any  force  polygon 
for  the  forces  would  close;  hence  the  resultant  of  the  forces  is  a   couple 

(Art.  6).  Any  couple  whose  moment  equals  the  alge- 
braic sum  of  the  moments  of  the  given  forces  about 
i^bs.  any  point  may  be  regarded  as  the  resultant.  For  ex- 
ample, let  us  find  the  resultant  of  six  forces  acting  on 
a  drawing  board  4  by  4  feet,  as  shown  in  Fig.  36.  In- 
spection shows  that  the  algebraic  sums  of  the  x  and 
the  y  components  equal  zero,  and  so  the  resultant 
is  a  couple.  Taking  moments  about  the  lower  left- 
hand  corner,  we  get  2i/=  —60  X  1.414  —  20  X  1.414 

—  18X3—  12X  I  =  —179  foot-pounds,  and  this  is  also  the  value  of  the 
moment  of  the  resultant  couple. 

If  the  forces  consist  of  a  number  of  couples,  then  2Fj  =  2Fy  =  o,  and 
the  resultant  is  a  couple.      But  the  sum  of  the  moments  of  these  forces 


60  lbs. 


Art.  8 


27 


Fig.  37 


about  any  point  equals  the  sum  of  the  moments  of  the  couples;  hence  any 
couple  whose  moment  equals  the  sum  of  the  moments  of  the  given  couples 
may  be  regarded  as  the  resultant. 

8.   Moment  of  a  Force ;  Couples  * 

§  I.  Moment  about  a  Line.  —  Art.  5  relates  to  moments  of  forces  and 
to  couples  with  special  reference  to  coplanar  forces  and  couples.  In  some 
discussions  on  noncoplanar  forces  it  is  convenient  to  make  use  of  the  moment 
or  torque  of  a  force  with  respect  to  a  line;  this  is  defined  as  the  product 
of  the  component  of  the  force  perpendicular  to  the 
line  —  the  other  component  being  parallel  to  it  — 
and  the  distance  from  the  line  to  the  perpendicu- 
lar component,  or  to  the  force  (the  distances  being 
equal).  For  example,  let  F  (Fig.  37),  acting  on  a 
body  not  shown,  be  the  force,  and  LL'  the  line,  or 
axis  of  moments  as  it  is  called.  MN  is  any  plane 
perpendicular  to  the  axis,  represented  to  make  the 
figure  plain.  OACB  is  a  parallelogram  with  OC 
(representing  F)  as  diagonal,  and  sides  perpendicu- 
lar and  parallel  to  LL';  then  OA  and  OB  represent 
the  perpendicular  and  parallel  components  {Fi  and  F2)  referred  to;  and  the 
moment  of  F  about  LL'  is  the  product  of  Fi  and  PL. 

The  moment  of  a  force  with  respect  to  a  line  is  a  measure  of  the  tendency 
of  the  force  to  turn  the  body  to  which  the  force  is  applied  about  that  line. 
Thus,  when  the  force  is  parallel  to  the  line  the  moment  is  zero,  and  obvi- 
ously the  force  has  no  tendency  to  turn  the  body  about  the  line.  Again, 
when  the  force  is  perpendicular  to  the  line  the  moment  of  the  force  about 
the  line  equals  the  product  of  the  force  and  the  perpendicular  distance 
from  the  line  to  the  force,  and  it  is  shown  in  Art.  5  that  this  product  meas- 
ures the  tendency  of  the  force  to  turn  the  body  about  the  line.  Finally, 
when  the  force  F  is  not  parallel  nor  perpendicular  to  the  axis  of  moments 
(Fig.  37),  then  Fi  and  F2  together  are  equivalent  to  F,  and  their  combined 
turning  effect  equals  that  of  F.  But  F^  has  no  turning  effect;  therefore 
that  of  Fi  and  that  of  F  are  equal.  But  it  was  explained  that  Fi  X  LP 
(the  moment  of  Fi)  measures  the  turning  effect  of  Fi,  and  therefore  that 
product  also  measures  the  turning  effect  of  F. 

In  a  discussion  involving  moments  of  several  forces  about  a  line,  it  is 
generally  convenient  to  give  signs  to  the  moments  to  indicate  the  directions 
(clockwise  or  counter)  in  which  the  several  forces  would  turn  the  body 
about  the  line  if  it  were  free  to  rotate  about  that  line.  Whether  a  given 
rotation  is  clockwise  or  counter  depends  on  the  point  of  view;  in  a  par- 
ticular discussion  a  point  of  view  should  be  assumed  on  the  line  or  axis  of 
moments  and  outside  of  the  body,  so  that  all  rotations  would  be  seen  look- 
ing in  the  same  direction.      When  the  axis  of  moments  is  also  an  axis  of 

*  See  Art.  5  also. 


28 


Chap,  i 


coordinates,  then  it   is  customary  to  view  rotations  about   that   axis   from 
the  positive  end  of  the  coordinate  axis,  looking  in  the  negative  direction. 

Principle  of  Moments.  —  If  two  sets  of  forces  are  equivalent  (Art.  2), 
then  the  moment-sum  for  one  set  with  respect  to  any  line  equals  the  mo- 
ment-sum for  the  other  set  with  respect  to  the  same  line.  This  will  be 
granted  as  self-evident  by  most  students;  others  may  consider  this:  Let 
^1  and  6*2  denote  the  two  equivalent  sets  of  forces,  and  ^3  a  third  set  which 
would  balance  Si  and  hence  also  ^2.  Since  Sx  and  ^3  would  balance,  they 
would  not  turn  the  body  on  which  they  act  about  any  line;  hence  the 
moment-sums  for  Si  and  ^'3  with  respect  to  any  line  are  equal  in  value  but 
opposite  in  sign.  Likewise,  the  moment-sums  for  S2  and  S3  with  respect 
to  that  same  line  are  equal  in  value  and  opposite  in  sign.  The  moment- 
sums  for  ^i  and  S2  being  equal  to  the  same  thing,  are  therefore  equal. 

It  follows  from  the  preceding  that  the  moment-sum  for  any  set  of  forces 
with  respect  to  a  given  line  equals  the  moment  of  the  resultant  of  those 
forces  with  respect  to  the  same  line.  Also,  the  moment  of  a  force  about  any 
line  equals  the  moment-sum  of  its  components  with  respect  to  the  same 
line.  This  last  principle  suggests  a  second  method  for  computing  the 
moment  of  a  force  with  respect  to  a  line,  more  simple  than  the  first  method 
in  some  cases:  Resolve  the  force  into  three  rectangular  components,  one  of 
which  is  parallel  to  the  axis  of  moments;  compute  the  moment  of  each  of 
the  other  two  components  about  the  axis,  and  add  the  moments  alge- 
braically; this  sum  equals  the  moment  of  the  given  force.  For  an  example, 
we  compute  the  moment  of  a  100  pound  force  which  acts  upon  a  4  foot 

cube  as  shown  in  Fig.  38,  with  respect  to  those 
edges  marked  X,  F,  and  Z.  The  x,  y,  and  z  com- 
ponents of  the  force  are  37.2,  74.2,  and  55.7  pounds 
respectively  (see  Art.  4);  these  components  must 
be  concurrent  with  the  given  force.  Taking  A  as 
the  point  of  concurrence,  the  moments  are  com- 
puted as  follows:  -74-2  X  4  +  55-7  X  4  =  "74; 
-37.2  X  4  -  55.7  X  ^  =  -260;  and  '37.2  X  4  + 
74.2  X  2  =  297  foot-pounds.  With  point  of  con- 
currence taken  at  B  or  at  any  other  point  in  AB, 
the  same  result  would  be  obtained  for  the  moment. 
§  2.  Couples  (see  also  Art.  5).  — Two  couples  whose  planes  are  parallel 
and  whose  moments,  or  torques,  and  senses  are  the  same  are  equivalent. 
Proof  of  this  proposition  for  coplanar  couples  is  given  in  Art.  5;  proof  for 
noncoplanar  couples  follows.  Let  Pi  and  P2  (Fig.  39)  be  the  forces  of  one 
couple,  Qi  and  Q2  (not  shown)  the  forces  of  the  other,  and  p  and  q  the  arms 
of  the  couples  respectively;  then  by  supposition  Pp  =  Qq.  According  to 
Art.  5,  the  (2  couple  can  be  replaced  by  a  couple  in  its  own  plane  provided 
that  the  moment  and  sense  of  the  new  couple  equals  that  of  the  Q  couple. 
Let  Si  and  ^2  be  the  forces  of  that  replacing  couple,  Si  and  ^2  being  chosen 


100  >b5 


■37.2 


Art.  8 


29 


Fig.  39 


parallel  and  equal  to  Pi  and  P^;  then  the  arm  ab  of  the  S  couple  equals  />, 
and  abed  is  a  parallelogram.  We  now  show  that  the  P  couple  would  balance 
the  reversed  S  couple;  it  will  follow  that  the  P  and  6*  couples  are  equiva- 
lent, and  hence  also  the  P  and  Q  couples.  The  resultant  R'  of  P]  and 
—52  {Si  reversed)  equals  the  resultant  R"  of  P2 
and  —S\  (Si  reversed),  and  R'  and  R"  are  parallel 
and  opposite  in  sense.  Moreover,  R'  lies  midway 
between  Pi  and  52,  and  R"  lies  midway  between 
P2  and  Si;  therefore  each  resultant  acts  through 
the  center  of  the  parallelogram  abed,  and  hence 
they  are  colinear.  The '  resultants  therefore  bal- 
ance, and  hence  the  four  forces  Pi,  P2,  —Si,  —S2 
do  also.     Therefore,  etc. 

The  resultant  of  any  number  of  couples  is  a  eouple.  Proofs  of  this  prop- 
osition for  the  case  of  coplanar  couples  are  given  in  Arts.  6  and  7.  For 
the  case  of  noncoplanar  parallel  couples:  The  given  couples  can  be  re- 
placed by  equivalent  ones  respectively,  all  in  some  one  plane;  the  result- 
ant of  these  is  a  couple,  and  hence  the  resultant  of  the  given  ones  is  also  a 
couple.  For  the  case  of  nonparallel  couples:  Imagine  each  of  the  two 
couples  to  be  replaced  by  an  equivalent  couple,  and  let  the  four  forces  of 
the  replacing  couples  be  equal;  furthermore,  imagine  the  two  new  couples 
so  placed  (in  their  respective  planes)  that  a  force  of  one  couple  will  balance 

a  force  of  the  other.  See  Fig.  40  (perspec- 
tive), which  shows  the  two  replacing  couples, 
there  marked  P1P2  and  P3P4;  ex  is  the  angle 
between  the  planes  of  the  couples.  Since 
P2  and  P4  balance,  Pi  and  P3,  constituting  a 
couple,  are  equivalent  to  Pi,  P2,  P3  and  P4 
and  hence  to  the  two  original  couples. 

The  resultant  of  any  coplanar  or  parallel 
couples  can  be  determined  very  simply;  the 
resultant  is  any  couple  parallel  to  the  given 
couples,  its  moment  being  equal  to  the  alge- 
braic sum  of  the  moments  of  the  given  couples.  The  resultant  of  nonparallel 
couples  can  be  determined  best  from  their  vectors*  by  means  of  this  proposi- 
tion,—  The  vector  of  the  resultant  of  any  number  of  couples  equals  the  sum 
of  the  vectors  of  those  couples.  Proof:  Consider  first  two  couples,  say  the 
two  whose  resultant  was  found  in  the  preceding  paragraph.  Let  ABC  (Fig, 
41)  be  an  end  view  of  Fig.  40,  looking  along  the  line  AA']  that  is,  ABC  of 

*  The  vector  of  a  given  couple  is  perpendicular  to  the  plane  of  the  couple  (exact  posi- 
tion of  vector  immaterial);  its  length  is  equal  to  the  moment  of  the  couple  according 
to  some  scale  understood;  and  its  sense  agrees  with  the  sense  (rotation)  of  the  couple 
according  to  some  rule  of  agreement,  as  for  example  the  following:  Imagine  the  vector 
to  be  a  right-handed  screw  turning  with  the  couple;  then  the  arrowhead  on  the  vector 
must  point  in  the  direction  in  which  the  screw  advances. 


Fig.  40 


3©  Chap,  i 

Fig.  41  is  ABC  of  Fig.  40  in  true  proportions.  Then  AM  (perpendicular  to 
AB),  AN  (perpendicular  to  AC),  and  AO  (perpendicular  to  BC)  are  respec- 
tively the  vectors  of  the  two  given  couples  and  their  resultant,  provided  that 
the  lengths  of  the  vectors  are  proportional  to  the  moments  of  the  couples  Ffi, 
Ffz  and  Ff;  let  the  lengths  be  in  that  proportion.  Vector  AO  is  the  sum 
of  the  vectors  AM  and  AN,  provided  that  OMAN  is  a  parallelogram;  we 
now  show  that  it  is  a  parallelogram.  Angle  MAO  =  13;  since  in  the  tri- 
angle MAO  and  ABC  two  sides  are  proportional  each  to  each  and  the  in- 
cluded angles  are  equal,  the  triangles  are  similar;  it  follows  that  OM  is 
perpendicular  to  AC,  or  parallel  to  AN.  From  similar  reasoning,  it  fol- 
lows that  ON  is  perpendicular  to  AB,  or  parallel  to  AM.  Hence  OMAN 
is  a  parallelogram.  Obviously,  if  the  proposition  holds  for  two  couples,  it 
holds  for  any  number. 

Composition  of  three  couples  whose  planes  are  mutually  at  right  angles 
is  an  important  special  case.  We  take  the  three  planes  as  coordinate 
planes,  and  call  the  couples  whose  planes  are  perpendicular  to  the  x,  y,  and 
0  axes  Cx,  Cy,  and  Cg  respectively,  their  vectors  Vx,  Vy  and  Vz,  and  the  re- 
sultant couple  C  and  its  vector  v.    Then  v  =  {v^^  +  Vy"^  +  ^z^)^;  hence 

Also,  if  01,  4>2,  and  ^3  denote  the  direction  angles  of  v,  then  cos  </)i  =  v^/v^ 
cos  02  =  ^vA,  and  cos  03  =  Vz/v\  hence 

cos  01  =  Cx/C,      cos  02  =  Cy/C,      cos  03  =  Cz/C. 

It  follows  from  the  preceding  that  a  couple  may  be  equivalent  to  two  or 
more  couples,  which  are  therefore  components  of  that  couple;  also,  to  re- 
solve a  couple  we  have  only  to  resolve  its  vector,  the  component  vectors 
being  the  vectors  of  the  component  couples.  The  resolution  of  a  couple 
into  three  components  whose  planes  are  mutually  at  right  angles  is  an  im- 
portant special  case.  Let  C  be  the  couple  to  be  resolved  and  v  its  vector, 
and  denote  the  direction  angles  of  the  vector  by  a,  /3,  and  7,  the  coordi- 
nate planes  having  been  taken  to  coincide  with  the  planes  of  the  desired 
component  couples.  Let  Cx,  Cy,  and  Cz  denote  the  component  couples, 
which  are  perpendicular  to  the  x,  y  and  z  axes  respectively,  and  Vx,  Vy  and 
Vz  the  corresponding  vectors.     Then  Vx=  v  cos  a,  Vy  ^  v  cos  /3,  and  Vz  =  v  cos  7; 

hence, 

Cx  =  C  cos  a,      Cy  =  C  cos  iS,      Cz  =  C  cos  7. 

9.   Noncoplanar  Nonconcurrent  Forces 

§  I.  Parallel  Forces. — It  is  shown  in  Art.  7  that  the  resultant  of  any  two 
parallel  forces  is  parallel  to  those  forces,  and  that  its  magnitude  and  sense 
are  given  by  the  algebraic  sum  of  the  forces,  the  sense  being  given  by  the 
sign  of  the  sum.  It  follows  that  the  resultant  of  any  number  of  parallel 
forces,  not  coplanar  necessarily,  is  parallel  to  the  forces,  and  that  its  magni- 


Art.  9 


31 


tude  and  sense  are  given  by  the  algebraic  sum  of  the  forces  (all  forces  of  the 
same  sense  having  one  sign,  and  those  of  the  opposite  sense  having  the  oppo- 
site sign).  The  line  of  action  of  the  resultant  may  be  fixed  by  means  of  the 
arms  of  the  resultant  with  respect  to  two  rectangular  axes,  each  perpendicu- 
lar to  the  forces.  Such  arms  can  be  computed  readily  from  the  principle 
that  the  moment  of  the  resultant  about  any  axis  equals  the  algebraic  sum 
of  the  moments  of  the  forces  about  the  same  axis. 

For  an  example,  we  find  the  resultant  of  four  forces  which  referred  to  a 
set  of  rectangular  axes  are  described  as  follows:  They  are  parallel  to  the 
z-axis;    their  magnitudes  are  recorded  in  the  first  column  of  the  schedule 


F 

X 

V 

M. 

^fy 

+  40 

-30 
—  20 

+  60 

-4 
-3 

+6 

+  7 

+  2 
—  2 

+5 
-8 

+  80 
+  60 

—  100 

-480 

+160 
-  90 

+  120 
—420 

+  50 

-440 

—  230 

adjoining,  the  signs  indicating  the  senses  of  the  forces;  their  lines  of  action 
pierce  the  xy  plane  in  points  whose  coordinates  are  recorded  in  the  second 
and  third  columns.  The  algebraic  sum  of  the  forces  is  +50,  and  so  the 
magnitude  of  the  resultant  is  50  pounds,  and  it  acts  in  the  positive  z  di- 
rection. The  moments  of  the  forces  about  the  x  and  y  axes  are  recorded  in 
the  last  two  columns  respectively.  The  algebraic  sums  of  these  moments 
are  —440  and  —230  foot-pounds,  as  indicated.  These  sums  are  also  the 
values  of  the  moments  of  the  resultant  with  respect  to  those  axes;  hence 
the  resultant  (acting  in  the  positive  direction)  is  below  the  x  axis  and  to  the 
right  of  the  y  axis  at  distances  equal  to  440  -^  50  or  8.8,  and  230  -r-  50 
=  4.6  feet  respectively. 

If  the  algebraic  sum  of  the  forces  equals  zero,  then  their  resultant  is,  in 
general,  a  couple.  For  the  resultant  of  all  the  forces  but  one  is  a  single 
force  equal  and  opposite  to  that  one;  and,  in  general,  that  resultant  force 
and  the  omitted  force  would  not  be  colinear,  and  so  they  constitute  a 
couple  as  stated. 

§  2.  NoNPARALLEL  FORCES. — A  set  of  noncoplanar,  nonconcurrent,  non- 
parallel  forces  may  be  compounded,  in  general,  into  a  force  acting  through 
any  arbitrary  chosen  point  arid  a  couple.  Proof  follows:  As  explained  in 
Art.  5,  each  force  of  the  given  system  may  be  resolved  into  and  be  re- 
placed by  a  force  acting  through  the  chosen  point  and  a  couple.  Sup- 
posing such  a  replacement  made  for  each  given  force,  then  the  new  system 
consists  of  a  set  of  concurrent  forces  at  the  chosen  point  and  a  set  of 
couples;  but  the  resultant  of  the  concurrent  forces  is  a  single  force  acting 
through   the  chosen  point    (Art.  4),  and   the  resultant  of   the  couples   is  a 


32 


Chap,  i 


single  couple  (Art.  8).     This  force  and  couple  respectively  will  be  denoted 
by  R  and  C. 

We  now  show  in  detail  how  to  determine  R  and  C.  Let  Fi,  F2,  F3,  etc. 
(Fig.  42,  only  T^i  shown),  be  the  forces  of  the  given  system  acting  on  a 
body  not  shown;  O  the  point  through  which  R  is  to  pass;  and  OX,  OY 
and  OZ  any  convenient  axes  of  reference.  Let  Pi  and  Qi,  acting  at  0 
(Fig.  42),  be  equal  and  parallel  to  Fi;  similarily,  let  P2  and  Q2  (not  shown) 
act  at  O,  and  be  equal  and  parallel  to  F^;  etc.  Then  the  force  Pi  and  the 
couple  Pi^i  (Fig.  43)  are  equivalent  to  Pi  (Fig.  42);  the  force  P2  and  the 
couple  F2Q2  are  equivalent  to  P2;   etc.     Now  the  axial  components  of  Pi, 


Y 

y 

y 

A 

/ 

0 

L_ 

/ 
/ 

/ 

/ 

/ 

Fig.  42 


Fig.  43 


Fig.  44 


P2,  Pz,  etc.  (the  concurrent  forces),  are  respectively  equal  to  the  axial  com- 
ponents of  Pi,  P2,  P3,  etc.  (the  given  forces);  hence  if  SP^,  2P„  and  2Pa 
denote  the  algebraic  sums  of  the  x,  y,  and  z  components  of  the  given  forces, 
then  Px  =  2P:,,  Ry  =  ZFy,  and  P,  =  ZP,;  also 

P2=(2P,)2+(SP,)2+(2:f.)2.  (i) 

And  if  01,  02,  and  ^3  denote  the  direction  angles  of  R,  then 

cos  01  =  XF^/R,      cos  02  =  ^Fy/R,      cos  0^  =  ZF,/R.  (2) 

These  formulas  determine  P.  To  determine  C:  Imagine  it  resolved  into 
three  components  whose  planes  are  respectively  perpendicular  to  the  x, 
y,  and  z  axes  (Art.  8),  and  denote  the  components  arid  their  moments  by 
Cx,  Cy,  and  Cz  (Fig.  44).  Since  the  system  P,  d,  Cy,  and  Cz  is  equivalent  to 
the  given  system,  their  moments  about  any  line  are  equal ;  hence  Cx  —  wi/^, 
Cy  =  ^My,  and  C^  =  ^M^,  where  2i/x,  ^My,  and  XM^  denote  the  moment- 
sums  for  the  given  system  with  respect  to  the  x,  y,  and  2  axes  respectively. 
Also,  according  to  Art.  8, 

C2  =  (2ilf,)2  +  (ZMyy  +  (2M.)2;  (3) 

and  if  0i,  02  and  03  denote  the  direction  angles  of  the  vector  representing 
C,  then 

cos  01  =  2lf  ^/C,        cos  02  =  ^My/C,        cos  03  =  ZMr/C.  (4) 


Art.  9  33 

In  general,  R  and  C  may  be  compounded  into  two  noncoplanar  forces. 
For,  as  explained  in  Art.  8,  C  may  be  shifted  about  without  change  of 
efifect  if  only  the  direction  of  its  plane  be  unchanged;  assume  such  shift 
until  one  of  the  forces  of  C  intersects  R;  then  that  force  and  R  may  be 
compounded  into  a  single  force  R';  there  remain  R'  and  the  second  force 
of  C,  and  obviously  R'  and  that  force  are  not  coplanar.  These  two  cannot 
be  compounded;  they  are  the  simplest  set  equivalent  to  the  given  system, 
and  therefore  constitute  the  resultant  of  the  given  system.  If  the  plane  of 
C  happens  to  be  parallel  to  R,  then  C  and  R  can  be  compounded  into  a 
single  force,  and  the  resultant  of  the  given  system  is  a  single  force.  For 
shifting  C  about  until  C  and  R  become  coplanar,  then  they  may  be  com- 
pounded readily  into  a  single  force  (Art.  5). 

In  general,  the  system  of  forces  has  a  torque  about  every  line  through  O. 
There  is  one  line  which  is  of  prime  importance,  the  line  about  which  the 
torque  is  greatest.  The  torque  of  the  forces  about  that  line  is  called  the  or 
resultant  torque  of  the  system  (for  the  chosen  point  0).  Since  R  has  no 
moment  about  a  line  through  O,  the  torque  of  the  system  about  any  such  line 
equals  the  torque  of  C  about  that  line.  But  the  torque  of  C  is  greatest  about 
a  line  perpendicular  to  the  plane  of  C;  this  is  the  important  line  mentioned. 
The  direction  of  this  line  is  given  by  equations  (4),  and  the  resultant  torque 
of  the  system  by  equations  (3).  The  system  of  forces  has  no  torque  about  a 
line  through  0  parallel  to  the  plane  of  C,  (perpendicular  to  the  line  or  axis  of 
resultant  torque)  since  R  and  C  have  no  torque  about  such  line. 


Ceiling 

■////////W////////////////, 


CHAPTER   II 

FORCES  IN  EQUILIBRIUM 

10.   Principles  of  Equilibrium 

§  I.  General  Conditions  of  Equilibrium. — It  is  convenient  in  some 
discussions  to  distinguish  forces  as  "external"  or  "internal,"  meaning  by 
external  force  one  which  is  exerted  on  the  body  under  discussion  by  some 
other  body,  and  by  internal  force  one  which  is  exerted  on  a  part  of  the  body 
under  discussion  by  another  part.  (The  word  body  is  used  here  in  a  broad 
sense  to  denote  any  definite  portion  of  matter,  as  a  locomotive,  a  bridge,  the 
steam  in  a  boiler,  the  water  in  a  pond,  etc.)     For  illustration,  consider  the 

crude  crane  in  Fig.  45.  It  consists  of  three 
main  members  {AB,  CD  and  DE),  a  pulley,  a 
winding  drum  and  a  hoisting  chain;  it  is  sup- 
ported at  A  (ceiling)  and  at  B  (floor).  The 
external  forces  acting  on  the  crane  consist  of 
the  weight  of  all  the  parts  (exerted  by  the 
earth),  the  pull  down  on  the  hook  (exerted  by 
/I  Tx^/  the  load),  the  supporting  force  at  A   (exerted 

'^.\y  by  the  ceiling),  and  the  supporting  force  at  B 

(exerted  by  the  floor).  The  members  exert 
forces  upon  each  other  where  they  come  to- 
gether, but  these  are  internal  forces  with  ref- 
erence to  the  whole  crane.  With  reference  to 
the  crane  post  AB,  the  external  forces  are  its  weight,  the  supporting  force  at 
A,  that  at  B,  the  pressures  on  it  at  E,  C,  and  the  drum.  All  these  are 
exerted  on  the  post  by  something  else,  and  so  are  properly  called  external 
forces.  Any  two  adjacent  portions  of  the  post,  as  the  upper  and  lower 
halves,  exert  forces  on  each  other,  and  these  forces  are  internal  with  refer- 
ence to  the  post. 

All  the  external  forces  acting  on  a  body  at  rest  constitute  a  balanced 
system,  and  such  system  is  said  to  be  in  eguilibrium.  Obviously,  the  re- 
sultant of  such  a  system  is  nil,  and  this  fact  is  sometimes  called  the  general 
condition  of  equilibrium  for  any  kind  of  a  force  system.  The  general  con- 
dition implies  subordinate  conditions;  thus,  for  any  system  whatever, 

(A)  the  algebraic  sum  of  the  (rectangular)  components  of  all  the  forces  along 

any  line  equals  zero,  and 
{B)  the  algebraic  sum  of  the  moments  of  all  the  forces  about  any  line  equals 


y/Mw//////////////////, 

Fig.  45 


zero. 


34 


Art.  io  35 

By  means  of  (A)  and  (B)  we  can  write  many  equations  for  any  system  in 
equilibrium.  Thus,  for  a  coplanar  concurrent  system,  (A)  gives  ZFx  =  o, 
ZFy  =  o,  ZFu  =  o,  etc.,  where  x,  y,  u,  etc.,  are  axes  of  resolution;  and 
(B)  gives  2ilfa  =  o,  I,Mb  =  o,  2Mc  =  o,  etc.,  where  a,  b,  c,  etc.,  are  origins 
of  moments  in  the  plane  of  the  forces.  Not  all  of  such  equilibrium  equa- 
tions are  independent,  however;  that  is,  certain  ones  follow  from  the  others. 
Thus,  if  2Fx  =  o  for  any  coplanar  concurrent  system,  then  2F„  does 
not  necessarily  equal  zero,  but  if  also  ^Fy  =  o,  then  the  resultant  equals 
zero,  and  it  follows  that  SFu  =  o.  That  is,  2^Fx  =  o  and  SF^  =  o  are  two 
independent  equations,  but  any  third  similar  equation  (as  ZFu  =  o)  is  not 
independent  of  them.  The  independent  equations  or  conditions  of  equi- 
librium for  any  particular  kind  of  force  system  are  such  as  are  necessary 
and  sufficient  to  insure  a  vanishing  resultant.  We  will  now  deduce  these 
independent  conditions  of  equilibrium  for  the  various  classes  or  kinds  of 
force  systems. 

(i)  Colinear  Forces. — There  is  one  condition  of  equilibrium.      It  can  be 
stated  in  several  forms;  namely, 

(i)  2/^  =  0      or  (2)  llMa  =  o. 

Form  (i)  states  that  the  algebraic  sum  of  the  forces  equals  zero;  (2)  that 
the  algebraic  sum  of  the  moments  of  all  the  forces  about  any  point  (not  on 
their  common  line  of  action)  equals  zero.  On  the  graphical  basis,  the  condi- 
tion of  equilibrium  is  that  the  force  polygon  for  the  forces  (degenerated  into 
a  straight  line  in  this  case)  is  a  closed  one.  For  if  ZF  =  o,  or  SM  =  o,  or 
the  force  polygon  closes,  then  there  is  no  resultant. 

(ii)  Coplanar  Concurrent  Forces. — There  are  two  independent  algebraic  con- 
ditions of  equilibrium.     They  can  be  expressed  in  three  forms;  namely, 

(i)  2Fx  =  2F„  =  o,      (2)  2/?  =  ZMa  =  o,      or       (3)  SMa  =  2^6  =  o. 

Form  (i)  states  that  the  algebraic  sums  of  the  components  of  the  forces 
along  two  lines  x  and  y  (in  the  plane  of  the  forces)  equal  zero;  (2)  that  the 
algebraic  sum  of  the  components  of  the  forces  along  any  line  (as  x),  and  the 
algebraic  sum  of  the  moments  of  all  the  forces  about  any  point,  each  equal 
zero  (the  point  a  to  be  in  the  plane  of  the  forces,  and  the  line  joining  a  and 
0,  their  point  of  concurrence,  to  be  inclined  to  the  x  axis);  and  (3)  that 
the  algebraic  sums  of  the  moments  of  all  the  forces  about  two  points  (not 
colinear  with  the  point  of  concurrence  of  the  forces)  equal  zero.  For  in 
any  case  the  resultant  is  zero,  as  will  be  seen  from  this:  (i)  According  to 
Art.  4,  the  resultant  of  the  system,  if  there  is  one,  is  a  single  force  R,  given 
hy  R==  V(2F,)2-f  (2F„)2;  and  hence  if  2Fx  =  o  and  2F„  =  o,  R  must 
equal  zero.  (2)  If  ^Fx  =  o,  then  the  resultant,  if  there  is  one,  must  be 
perpendicular  to  the  x  axis;  and  if  2Ma  =  o,  then  the  moment  of  R  about 
a  equals  zero,  which  requires  that  R  ^  o.  (3)  The  resultant,  if  there  is  one, 
must  pass  through  the  point  of  concurrence  0  of  the  given  forces;  if  2Mo=  o 


36  Chap,  n 

then  R  must  pass  through  a  also;  if  ZMa  =  o,  then  R  must  equal  zero,  b 
not  being  on  Oa. 

The  graphical  condition  of  equilibrium  is  that  the  force  polygon  for  the 
forces  closes.     For,  if  it  does  close,  then  there  is  no  resultant. 

(iii)  Co  planar  Nonconcur  rent  Parallel  Forces. — There  are  two  independent 

algebraic  conditions  of  equilibrium.     They  can  be  expressed  in  two  forms; 

namely, 

(i)  S/?  =  SM  =  o      or      (2)  -^Ma  =  ^Mb  =  o 

Form  (i)  states  that  the  algebraic  sum  of  the  forces  and  the  algebraic  sum 
of  the  moments  of  the  forces  about  any  point  (in  the  plane  of  the  forces)  equal 
zero;  (2)  that  the  algebraic  sums  of  the  moments  of  the  forces  about  two 
points  equal  zero,  the  line  joining  the  origins  not  to  be  parallel  to  the  forces. 
For  either  set  of  conditions  is  necessary  and  sufficient  to  make  the  result- 
ant zero,  as  may  be  shown  thus :  In  Art.  7  it  is  shown  that  the  resultant, 
if  there  is  one,  is  a  single  force  or  a  couple.  And  (i),  if  2F  =  o,  then  the 
resultant  is  not  a  force,  and  if  2M  =  o,  then  it  is  not  a  couple;  and  hence 
there  is  no  resultant.  (2)  If  Sil/a  =  o,  the  resultant  is  not  a  couple  but  a 
force,  which  passes  through  a;  if  also  ZMb  =  o,  then  the  moment  of  the 
resultant  force  about  b  must  be  zero,  and  that  requires  that  the  force  equals 
zero. 

There  are  two  graphical  conditions  of  equilibrium,  namely,  a  force  and 
a  string  polygon  for  the  forces  must  close.  For  if  a  force  polygon  closes, 
then  the  resultant,  if  there  is  one,  is  a  couple;  if  a  string  polygon  closes,  then 
the  resultant  is  not  a  couple. 

(iv)  Coplanar  Nonconcurrent  Nonparallel  Forces. — There  are  three  inde- 
pendent algebraic  conditions  of  equilibrium.  They  can  be  stated  in  three 
forms;  namely, 

(i)    2/^x  =  ^Fy  =  2Ma  =  o; 

(2)    Zi^x  =  2M„  =  ^Mb  =  o; 

and  (3)    SMa  =  2Mb  =  ^Mc  =  o. 

Form  (i)  states  that  the  algebraic  sums  of  the  components  of  all  the  forces 
along  two  lines  and  the  algebraic  sum  of  the  moments  of  the  forces  about 
any  point  equal  zero,  the  lines  and  points  to  be  in  the  plane  of  the  forces; 
(2)  that  the  algebraic  sums  of  the  components  of  the  forces  along  any  line  x 
and  the  algebraic  sums  of  the  moments  of  the  forces  about  two  points,  a  and 
b,  equal  zero,  the  line  x  and  that  joining  a  and  b  not  to  be  at  right  angles; 
and  (3)  that  the  algebraic  sums  of  the  moments  of  the  forces  about  three 
points,  a,  b,  and  c,  equal  zero,  the  points  not  to  be  colinear.  For  any  set  of 
these  conditions  is  necessary  and  just  sufficient  to  make  the  resultant  vanish 
as  may  be  shown,  thus:  The  resultant,  if  there  is  one,  is  a  single  force  or  a 
single  couple  (Art.  7).  And  (i)  if  ^Fx  =  2Fj,  =  o,  then  the  resultant  is  not 
force,  and  if  2M  =  o,  it  is  not  a  couple;  and  hence  there  is  no  resultant. 
(2)  If  2Fx  =  o,  the  resultant  is  a  force  R  perpendicular  to  the  x  axis  or  a 


Art.  io  37 

couple;  if  l^Ma  =  o,  it  is  not  a  couple,  but  a  force  passing  through  a  (and 
perpendicular  to  the  x  axis);  if  also  SMt  =  o,  then  the  moment  of  that 
force  about  b  must  equal  zero,  and  hence  the  force  must  equal  zero.  (3)  If 
2ifa  =  o,  the  resultant,  if  there  is  one,  is  not  a  couple  but  a  force  passing 
through  a;  if  liMb  =  o,  that  resultant  passes  through  b;  if  also  ZMc  =  o, 
then  the  resultant  force  must  equal  zero. 

There  are  two  graphical  conditions,  just  like  those  for  parallel  coplanar 
nonconcurrent  forces;  namely,  a  force  and  a  string  polygon  must  close. 
For  if  a  force  polygon  closes,  then  the  resultant,  if  there  is  one,  is  not  a 
force  but  a  couple;  if  a  string  polygon  closes,  then  the  resultant  is  not  a 
couple,  and  so  there  is  no  resultant  (see  Art.  6). 

(v)  Noncoplanar  Concurrent  Forces.  —  There  are  three  independent  alge- 
braic conditions  of  equilibrium.     The  convenient  form  is 

2Fx  =  ^Fy  =  2F,  =  o; 

that  is,  the  algebraic  sums  of  the  components  of  all  the  forces  along  three 
rectangular  axes,  x,  y,  and  z,  equal  zero.  For  as  shown  in  Art.  4,  the  resultant, 
if  there  is  one,  equals  V(SF^)2  +  (SFJ^  ^  {-^F.Y,  and  so  if  the  conditions 
stated  are  fulfilled  then  the  resultant  equals  zero. 

(vi)  Noncoplanar  Parallel  Forces.  There  are  three  independent  algebraic 
conditions  of  equilibrium.     There  are  two  convenient  forms;  namely, 

(i)    SF  =  SMi  =  Si/a  =  o,       and  (2)    2Mi  =  Silfa  =  ^Ms  =  o. 

Form  (i)  states  that  the  algebraic  sum  of  the  forces  and  the  algebraic  sums 
of  the  moment  of  the  forces  about  two  lines  perpendicular  to  the  forces  but 
not  parallel  to  each  other  equal  zero;  (2)  that  the  algebraic  sums  of  the  mo- 
ments about  three  coplanar  nonconcurrent  nonparallel  lines  perpendicular  to 
the  forces  equal  zero.  For  (i)  if  SF  =  o,  the  resultant  is  not  a  force;  if 
2Afi  =  o,  the  resultant  is  a  couple  whose  plane  is  parallel  to  the  first  line  or 
axis  of  moments  (and  to  the  forces) ;  and  if  '2M2  =  o,  then  the  plane  of  the 
couple  must  also  be  parallel  to  the  second  axis;  but  all  these  conditions 
of  parallelism  cannot  be  fulfilled  unless  the  two  forces  of  the  couple  are 
colinear,  in  which  case  the  two  forces  balance,  so  that  there  is  really  no  re- 
sultant. (2)  If  2^/1  =  2M2  =  o,  then  the  resultant  must  be  a  force  pass- 
ing through  the  intersection  of  lines  i  and  2;  if  Sil/a  =  o,  then  that  force 
must  equal  zero;   that  is,  the  three  conditions  make  the  resultant  vanish. 

(vii)  Noncoplanar  Nonconcurrent  Nonparallel  Forces.  —  There  are  six 
independent  algebraic  conditions  of  equilibrium,  namely, 

ZF^  =  -LFy  =  ZF,  =  ZM:c  =  ZMy  =  ^M,  =  o; 

that  is,  the  algebraic  sums  of  the  components  of  all  the  forces  along  three 
lines  and  the  algebraic  sums  of  the  moments  of  the  forces  about  three  non- 
coplanar axes  equal  zero.  (It  is  generally  most  convenient  to  take  the 
three  lines  and  the  three  axes  at  right  angles  to  each  other.)     For  the  result- 


Fig.  46 


28  Chap,  ii 

ant  of  the  system,  if  there  is  one,  is  always  reducible  to  a  single  force  and 
a  single  couple  (Art.  9) ;  if  SF^  =  2Fj,  =  SF^  =  o,  the  single  force  equals 
zero,  and  if  21f ^  =  2My  =  SM^  =  o,  then  the  couple  vanishes,  and  so  there 
is  no  resultant. 

If  every  force  in  the  given  system  (in  equilibrium)  be  represented  by  a 
vector,  and  all  these  vectors  be  projected  on  three  rectangular  coordinate 

planes,  then  the  three  sets  of  projections 
represent  three  force  systems,  and  each 
is  in  equilibrium  (proved  below).  In 
some  cases  it  may  be  more  convenient  to 
deal  with  these  projected  systems.  In 
general,  each  furnishes  three  conditions 
or  equations  of  equilibrium,  making  nine 
in  all;  but  there  are  duplicates  among  the 
nine,  and  only  six  are  independent.  To 
prove  the  foregoing,  let  F  (Fig.  46)  be 
one  of  the  forces  of  the  system  in  equilib- 
rium and  P  its  point  of  application  (on  a 
body  not  shown).  A,  B,  and  C  are  pro- 
jections of  the  vector  F  on  the  xy,  yz,  and  zx  planes  respectively.  Obviously, 
the  X  and  y  components  of  A  equal  Fx  and  Fy  respectively;  the  y  and  z  com- 
ponents of  B  equal  Fy  and  F^  respectively,  and  the  z  and  x  components  of 
C  equal  F^  and  Fx  respectively,  as  indicated.     Since  the  given  system  is  in 


and 


Now  2Fx  is  also  the  sum  of  the  x  components  of  the  ^-system;  ZFy  is  also 
the  sum  of  the  y  components  of  the  ^-system;  and  XiFyX—Fxy)  is  also  the 
sum  of  the  moments  of  the  A  forces  about  0.  Hence  (i),  (2),  and  (6)  are 
conditions  which  assert  the  equilibrium  of  the  yl -system.  For  similar  rea- 
sons (2),  (3),  and  (4)  assert  the  equilibrium  of  the  ^-system  and  (i),  (3), 
(5)  assert  the  equilibrium  of  the  C-system. 

§  2.  Special  Conditions  of  Equilibrium,  depending  on  number  of  forces 
in  the  system.  —  (i)  A  single  force  cannot  be  in  equilibrium.  (2)  If  two 
forces  are  in  equilibrium,  then  obviously  they  must  be  colinear,  equal,  and 
opposite.  (3)  If  three  forces  are  in  equilibrium,  then  they  must  be  coplanar, 
and  concurrent  or  parallel.  Proof:  Let  the  three  forces  be  called  Fi,  F2, 
and  7^3;  since  Fi  and  F2  balance  F3,  Fi  and  F2  have  a  single  force  resultant 
R  colinear  with  F3;  since  Fi  and  F2  have  a  resultant  colinear  with  F3,  they 
lie  in  a  plane  with  F3.  If  Fi  and  F2  are  concurrent,  then  R  is  concurrent 
with  them  and  hence  F3  also;  if  Fi  and  F2  are  parallel,  then  R  and  hence 
F3  is  parallel  to  them.     When  the  three  forces  are  concurrent,  then  each  is 


equilibrium, 

(i) 

^Fx  =  0, 

(4) 

ZMx  =  ^{Fzy  -  Fyz)  =  0, 

(2) 

ZFy     =     0, 

(5) 

ZMy  =  XiFxZ  -  F^x)  =  0, 

(3) 

2F,  =  0, 

(6) 

2M,  =  ^(Fyx  -  Fxy)  =  0. 

Art.  io 


39 


proportional  to  the  sine  of  either  angle  between  the  other  two  (Lami's 
theorem) ;  that  is, 

Fi  ^  F2  ^  Fs 

sin  a  =  sin  a"       sin  /3'  =  sin  /3"      sin  7'  =  sin  7" ' 

where  Fi,  F2,  and  F3  are  the  forces,  a  and  a"  the  angles  between  F2  and  F3, 
13'  and  ^"  those  between  Fi  and  /^s,  and  y'  and  7"  those  between  i^i  and  F2 
(see  Fig.  47).  For  it  follows  from  the  triangle  of  forces,  ABC  A  (in  which 
AB,  BC,  and  CD  represent  Fi,  F^,  and  F3  respectively),  that  ^5/sin5C^  = 
BC/dn  CAB  =  CA/sin  ABC.  But  BCA  =  a',  CAB  =  /3',  and  ABC  =  y';  also 
a'  and  a",  ^'  and  (S",  7"  and  7"are  supplementary.  Hence  sin  «'  =  a  sin  ", 
etc.,  etc.  When  the  three  forces  are  par- 
allel, then  the  two  outer  ones  act  in  the 
same  direction  and  the  middle  one  in  the 
opposite  direction,  and  the  moments  of 
any  two  of  the  forces  about  a  point  on 
the  third  are  equal  in  magnitude  and  op- 
posite in  sense,  or  sign.  (4)  When  four 
coplanar  forces  are  in  equilibrium,  then 
the  resultant  R  of  any  two  of  the  forces 

balances  the  other  two.  Hence,  (a)  if  the  first  two  are  concurrent  and  the 
second  two  also,  then  the  R  passes  through  the  two  points  of  concurrence; 
Q})  if  either  two  are  concurrent  and  the  other  two  parallel,  then  the  resultant 
R  of  the  first  pair  acts  through  the  point  of  concurrence  and  is  parallel  to 
the  second  pair;  (c)  if  all  four  forces  are  parallel,  then  R  is  parallel  to 
the  forces.  Principles  (a)  and  {b)  are  useful  in  graphical  analysis  of  four- 
force  systems, 

§  3.   Summary. — The  algebraic  conditions  of  equilibrium  explained  in  detail 
in  the  foregoing  are  brought  together  here  for  convenience  of  reference. 

Coplanar  Forces. 

Colinear,        SF  =  o;  or  Sif  =  o. 

Concurrent,  2F^  =  2/^i;  =  o;  or  SFx  =  ^M^  =  o;  or  llMa  =  ^Mb 
Parallel,  2F  =  2M  =  o;  or  2Ma  =  Zilf^  =  o. 

Nonconcurrent  nonparallel,  ^Fx  =  llFy  =  ZM  =  o;  or 

2Fx  =  ^Ma  =  ^Mb  =  o;  or  SM^  =  ^Mb  =  ZMc  =  o. 


Fig.  47 


=  o. 


Noncoplanar  Forces. 

Concurrent,  ZF^  =  ZFy  =  ZF^  =  o. 

Parallel,         SF  =  SMi  =  2M2  =  o;  or  Sil/i  =  SATz  =  2M3  =  o. 

Nonconcurrent  nonparallel,    SF^  =  ZFy  =  XF^  =  XMx  =  2Af„  =  2ilf ,  =  o. 

The  graphical  conditions  of  equilibrium  for  coplanar  systems:  for  concur- 
rent forces,  the  force  polygon  closes;  for  nonconcurrent  forces,  the  force  and 


40 


Chap,  n 


the  string  polygon  close.  There  are  graphical  conditions  of  equilibrium  for 
noncoplanar  forces,  but  their  usefulness  is  very  limited,  and  they  are  there- 
fore not  given  here. 

II.    Coplanar  Concurrent  Forces  in  Equilibrium 

§1.  The  general  principles  of  equilibrium  for  such  forces  are  explained 
in  Art.  lo  under  (ii).  We  now  show  how  to  apply  the  principles  in  two 
particular  problems. 

Typical  Problem  (i).  —  A  system  of  coplanar  concurrent  forces  is  in 
equilibrium,  and  all  the  forces  except  two  are  wholly  known;  the  lines  of 
action  of  these  are  known,  and  their  magnitudes  and  senses  are  to  be  de- 
termined. The  graphical  method  is  generally  the  simplest  for  solving  this 
problem;  but  if  there  are  only  three  forces  in  the  system,  or  if  the  angle 
between  the  two  unknown  forces  is  90  degrees,  then  the  algebraic  method 
is  simple. 

To  solve  graphically,  we  draw  a  force  polygon  for  all  the  forces,  and 
make  it  close  since  they  are  in  equilibrium;  in  doing  so  the  desired  un- 
knowns will  be  determined.  For  example,  consider  the  forces  acting  on 
the  pin  0  of  the  bridge  truss   partially   represented  in   Fig.  48.     (A  pin 


passes  through  holes  in  the  members,  OF,  OG,  OH,  and  OJ,  thus  fastening 
them  together  at  0.)  There  are  four  forces  acting  on  this  pin,  one  exerted 
by  each  member  named,  and  they  constitute  a  system  in  equilibrium. 
(Strictly,  there  is  a  fifth  force  in  the  system,  the  weight  of  the  pin,  but 
that  is  small  compared  to  the  others  and  is  negligible.)  These  four  forces 
are  coplanar  and  concurrent.  We  assume  that  they  act  in  the  directions 
of  the  members  respectively  (generally  not  far  from  the  fact)  as  shown; 
furthermore,  we  will  suppose  that  the  magnitudes  and  directions  of  two  of 
the  forces  have  been  detennined  somehow.  Now  to  determine  the  other 
two,  P  and  Q,  completely:  We  draw  AB  to  represent  the  80  ton  force  accord- 
ing to  some  convenient  scale;  and  BC  to  represent  20  tons;  then  from  C,  a 
line  parallel  to  Q,  and  from  A,  a  line  parallel  to  P,  and  mark  their  intersec- 
tion D.  Then  CD  and  DA  represent  the  magnitudes  Q  and  P  respectively; 
and,  since  the  arrowheads  in  the  closed  vector  polygon  must  be  confluent  Q 
acts  in  the  direction  CD  and  P  in  the  direction  DA.  There  are  other  possible 
force  polygons,  each  giving  the  same  result  as  the  one  explained. 


Art.  II 


41 


20  tons 
Fig.  49 


To  solve  this  problem  algebraically  we  may  employ  any  one  of  the  three 
sets  of  equations  or  conditions  of  equilibrium  (Art.  10) ;  namely, 

2Fx  =  ^Fy  =  o,  2F^  =  ZMa  =  o,  or  XMa  =  SM^  =  o. 

Taking  the  first  set  and  assuming*  senses  for  P  and  Q  (Fig. 
49),  we  get 

2Fx  =  Q  cos  20°  4-  P  cos  40°  +  80  cos  40°  =  o,  and 

ZFy  =  —  20  +  (J  sin  20°  —  P  sin  40°  +  80  sin  40°  =  o; 

solving  these  equations  simultaneously  for  P  and  Q,  we  get  P  =  10.04  and 

(2  =-73-3  tons. 

When    the   system   is  a   three-force   system,  then  the  special   condition, 

Fi/sin  a  =  /^2/sin  13  =  Fa/sin  y  (Art.  10),  is,  in  general,  the  simplest  to  apply. 

(Fi,  F2  and  F3  denote  the  forces,  and  a  either  angle  between  Fo  and  F3,  /3 

either  angle  between  F3  and  Fi,  and  7  either  angle  between  Fi  and  F2.) 

To  illustrate,  we  discuss  the  forces  acting  upon  a  cylin- 
der which  lies  in  a  trough  formed  by  two  smooth  f  in- 
clined planes  (Fig.  50).  There  are  three  forces  acting  on 
the  cylinder;  namely,  its  own  weight  (100  pounds),  and 
the  two  supporting  forces  Fi  and  F2.  Since  the  planes 
are  smooth  Fi  and  F2  act  normally,  and  hence  through 
the  center  of  the  cylinder  as  shown.  It  follows  from  the 
geometry  of  the  figure  that  the  acute  angle  between  Fi 
and  W  =  40°,  that  between  F2  and  W  =  80°,  and  that 

between  Fi  and  F2  =  60°;  hence  Fi/sin  80°  =  F2/sin  40°  =  loo/sin  60°,  or  F, 

=  1 13.7  and  F2  =74.2  pounds. 

*  Whenever  a  force  whose  sense  is  unknown  is  to  be  entered  in  a  resolution  or  moment 
equation,  a  sense  should  be  assumed  for  that  force  and  adhered  to  in  the  solution  of  the 
equation.  The  correct  sense  is  indicated  by  the  sign  of  the  computed  value  of  that  force; 
a  positive  sign  indicates  that  the  sense  assumed  is  correct  and  a  negative  sign  that  the 
sense  assumed  is  wrong.  Senses  found  to  be  wrong  are  corrected  in  the  figures  of  the 
book,  by  a  short  line  across  the  assumed  arrowhead  (Fig.  49). 

t  When  two  bodies  are  in  contact,  and  they  exert  forces  upon  each  other  (equal  and 
opposite),  the  forces  are,  in  general,  inclined  to  the  surface  of  contact,  assumed  plane  for 
the  moment.  The  components  of  either  of  the  forces  men- 
tioned along  and  perpendicular  to  the  surface  of  contact  are 
called  friction  and  normal  pressure  respectively.  Fig.  51  fur- 
nishes the  simplest  illustration;  it  represents  a  heavy  body  A 
supported  by  a  rough  surface  B,  and  subjected  to  a  push  P. 
The  surface  B  exerts  a  force  R  on  A  (inclined  as  shown),  and 
the  horizontal  and  vertical  components  of  R  are  the  friction 
and  the  normal  pressure  exerted  by  B  on  A.  Obviously,  this  friction  is  the  resistance  which 
B  offers  to  the  tendency  of  A  to  slide  over  B.  So  long  as  there  is  only  tendency  to 
sliding,  this  friction  equals  the  push  P.  Experience  has  shown  that  the  friction  is  a  maxi- 
mum just  as  sliding  impends,  and  also  that  the  smoother  the  surfaces  of  contact,  the 
smaller  is  the  force  required  to  cause  sliding,  and  hence  the  smaller  this  maximum  resist- 
ance to  sliding.    We  are  thus  led  to  the  conception  of  a  perfectly  smooth  surface  as  one 


W 


■'W/////I 


V^/ 


TrmrmrnTTTmnrrm 


Fig.  51 


42 


Chap,  u 


nWNWWNWvww. 


1fi\ 


Uc 


Typical  Problem  (ii).  A  system  of  coplanar  concurrent  forces  is  in 
equilibrium  and  all  except  one  are  wholly  known;  the  magnitude  and  direc- 
tion of  this  one  are  required.  To  solve  this  problem  we  might  determine 
the  resultant  of  the  w^hoUy  known  forces;  this  resultant  reversed  is  the 
desired  force.  But  the  problem  may  also  be  solved  by  means  of  principles 
of  equilibrium,  that  is,  by  applying  the  appropriate  condi- 
i"  tions  of  equilibrium  to  the  entire  system  of  forces.  To 
illustrate,  we  determine  the  value  and  direction  of  the  ten- 
^  sion  in  the  cord*  (Fig.  52)  which  supports  a  ring  from 
\  which  a  body  W  is  suspended,  the  ring  being  subjected  to 
a  force  P  as  shown.  The  forces  acting  on  the  ring  are 
W  100  lbs    B  W,  P,  and  the  pull  of  the  long  cord  (equal  to  the  tension), 

Fig.  52  and  these  three  forces  are  in  equilibrium.     To  solve  graph- 

ically, we  draw  AB  to  represent  W,  and  BC  to  represent  P; 
then  CA  represents  the  desired  pull  or  tension.  To  solve  algebraically,  we 
call  the  desired  force  F  and  its  inclination  to  the  vertical  6.  Then,  using 
the  conditions  ZFx  =  o  and  ZFy  =  o,  we  get  20  cos  30°  —  F  sin  ^  =  o  and 
—  100  +  F  cos  6  -\-  20  sin  30°  =  o;   these 

solved     simultaneously     give     F  —  91.6    J''    A C B P^ 

pounds,  and  6  —  10°  54'.  yig.  53 

As  another  example,  we  determine  the 
force  which  the  inclined  plane  (Fig.  54)  exerts  on  the  body  A  when  it  is  sub- 
jected to  a  pull  P  =  20  pounds,  the  plane  being  so  rough  that  motion  does 
not  ensue.  The  weight  of  A  (100  pounds),  P,  and  the  re- 
action R  of  the  plane  are  in  equilibrium;  hence,  using  6  to 
denote  the  inclination  of  R  to  the  plane,  and  resolving  along 
the  plane  and  normal  to  it,  we  get 

20  —  100  sin  30  +  i?  cos  9  =  0,  and  i^  sin  ^  —  100  cos  30°  =  o. 

Solving  these  simultaneously,  we  get  R  =  91.7  pounds,  and 
Fig.  54  6  =  70°  53'. 


which  can  ofifer  no  frictional  resistance,  only  normal  reaction.  Such  a  surface  is  of 
course  ideal,  but  there  are  surfaces  which  are  nearly  perfectly  smooth.  For  brevity  we 
will  call  these  smooth,  and  those  whose  resistance  to  sliding  is  to  be  taken  into  account 
will  be  called  rough. 

If  the  surface  of  contact  between  two  bodies  is  curved,  then  we  speak  of  the  friction 
and  normal  pressure  at  any  elementary  portion  of  the  contact,  meaning  the  tangential 
and  normal  components  of  the  pressure  at  that  element.  If  the  contact  between  two 
bodies  is  small,  practically  a  point,  and  they  exert  forces  R  upon  each  other  there,  then 
normal  pressure  means  the  component  of  R  at  right  angles  to  the  plane  which  is  tangent 
to  the  surfaces  at  the  contact,  and  friction  means  the  component  along  that  plane.  If 
one  or  both  the  bodies  is  smooth,  then  any  pressure  exerted  between  the  two  at  any  point 
of  the  contact  is  directed  along  the  normal  there.  (For  fuller  discussion  of  friction  see 
Chapter  IV.) 

*  "  Tension  in  a  cord  "  refers  to  the  forces  which  two  parts  of  a  taut  cord  exert  upon  each 
other.    Suppose  thai;  AB  (Fig.  53)  is  a  cord  subjected  to  equal  pulls  at  its  ends,  and  imagine  a 


Art.  II 


43 


§  2.  Many  machines  and  other  devices  consist  of  parts  (members)  more 
or  less  intimately  connected,  and,  in  general,  these  parts  exert  forces  upon 
each  other  when  the  machine  is  in  service.  To  determine  these  forces 
seems  a  compUcated  problem  lo  most  beginners.  And  yet  in  many  in- 
stances the  whole  problem  can  be  resolved  into  several  simpler  ones,  often 
like  typical  problem  (i),  which  may  be  solved  in  turn  and  thus  furnish 
values  of  the  desired  forces.  In  this  connection  it  will  be  convenient  to 
designate  a  member  of  any  device  as  a  one-force  piece  if  only  one  force 
acts  upon  it;  as  a  two-force  piece  if  only  two  forces  act  upon  it;  etc. 
Obviously,  a  one-force  piece  cannot  be  at  rest.  If  a  two-force  piece  is  at 
rest,  then  the  two  forces  acting  upon  it  must  be  equal,  opposite,  and  colinear; 
each  force  acts  in  the  line  joining  their  points  of  application,  and  the  re- 
actions which  the  piece  exerts  (upon  the  members  which  act  upon  it)  also 
act  along  the  same  line.*  If  a  three-force  piece  is  at  rest,  then  the  three 
forces  are  coplanar,  and  concurrent  or  parallel  (Art.  lo,  §  2).  If  a  four-force 
piece  is  at  rest,  then  the  resultant  of  any  pair  of  the  four  balances  the  other 
pair.  We  now  illustrate  how  to  resolve  the  apparently  diflficult  problem  into 
several  simpler  ones. 

Example.  The  crab-tongs  represented  in  Fig.  55  consist  of  six  pieces 
fastened  together  by  pins  B,B' ,  C,C',  and  D\   angle  ABC  =  100  degrees, 


Fig.  55 

AB  =  I    foot,   BC  =  I    foot   9    inches,    CD  =  i    foot,    and   BB'  =  3    feet. 

Required  the  forces  which  act  on  each  piece  when    the  tongs  suspend  a 

stone  W  whose  weight  =  1000  pounds,  and  width  A  A'  =  i  foot  6  inches. 

Apparently,  the  trigonometric  relations  between  the  parts  are  not  simple; 

so  we  will  solve  graphically,  and  first  we  draw  (or  lay  out)   the  tongs  to 

scale.     Obviously,  the  supporting  force  at  E  equals  1000  pounds  (weight  of 

plane  of  separation  at  any  place  C  between  the  ends  of  the  cord.  Since  the  part  AC  is  in 
equilibrium,  there  is  a  force  acting  upon  it  at  its  right  end  equal  and  opposite  to  P';  this 
force  is  exerted  by  the  part  BC.  Similarly,  there  is  a  force  acting  upon  BC  at  its  left  end 
equal  and  opposite  to  P";  this  force  i.s  e.xerted  by  the  part  AC.  These  two  equal  and 
opposite  forces  at  C  hold  the  parts  AC  and  BC  together.  By  magnitude  of  the  tension 
is  meant  the  magnitude  of  either  of  the  forces. 

*  Action  and  reaction  are  equal,  opposite,  and  colinear  if  they  are  concentrated.  This 
is  a  brief  statement  of  Newton's  Third  Law  of  Motion,  and  it  means  that  when  one  body 
exerts  a  force  upon  another  body  then  the  latter  also  exerts  one  on  the  former,  and  the 
two  forces  are  equal  in  magnitude  and  opposite  in  direction.  By  action  is  meant  either 
of  these  two  forces  and  by  reaction  the  other  one. 


44  Chap,  n 

tongs  neglected).  The  pin  D  is  acted  upon  by  DE,  DC  and  DC,  and, 
since  each  of  these  is  a  two-force  piece,  the  forces  upon  the  pin  act  along 
DE,  DC,  and  DC,  as  shown  at  center.  The  first  force  equals  looo  pounds 
and  acts  upwards;  determination  of  the  other  two  presents  typical  prob- 
lem (i).  So  we  draw  MN  to  represent  the  looo  pound  force,  and  from  M 
and  N  lines  parallel  to  the  other  two,  thus  fixing  0;  then  NO  and  OM  rep- 
resent the  magnitudes  of  the  two  forces  (620  pounds).  It  follows  that  DC 
and  DC  are  subjected  to  end  pushes  or  compressions  of  620  pounds.  CBA 
is  a  three-force  piece,  the  forces  being  applied  at  C,  B  and  A.  The  first 
acts  parallel  to  CD  as  shown  and  equals  620  pounds;  the  second  is  exerted 
by  the  two-force  piece  BB',  and  hence  acts  along  BB'',  and  the  third  must  be 
concurrent  with  the  first  two  and  so  acts  along  the  straight  line  through  A. 
Determination  of  the  two  unknown  forces  presents  typical  problem  (i).  So 
we  draw  PQ  to  represent  the  620  pound  force,  and  lines  from  P  and  Q 
parallel  to  the  other  two,  thus  fixing  R;  then  QR  represents  the  force  at  A 
(950  pounds),  and  RP  that  at  B  (1315  pounds).  It  follows  that  the  piece 
BB'  is  subjected  to  end  pulls  of  13 15  pounds. 

12.   Coplanar  Parallel  Forces  in  Equilibrium 

§  I.  Principles  of  equilibrium  for  a  system  of  forces  of  this  kind  are  de- 
veloped in  Art.  10  under  (iii) ;  we  now  show  how  to  apply  them  to  a  common 
problem.     (For  typical  problems  i  and  ii  see  Art.  11.) 

Typical  Problem  (iii).  A  system  of  coplanar  parallel  forces  is  in  equilib- 
rium, and  all  the  forces  except  two  are  wholly  known;  the  lines  of  action 
of  these  two  are  known  and  their  magnitudes  and  senses  are  required. 

The  algebraic  method  is  the  better  one,  by  far,  for  solving  the  problem. 
There  are  two  sets  of  conditions  of  equilibrium  available;  namely,  (i) 
27^  =  1,M  =  o,  that  is,  the  algebraic  sum  of  the  forces  and  the  algebraic 
sum  of  the  moments  of  the  forces  each  equal  zero;  and  (2)  Sifo  =  'LMb  =  c, 
that  is,  the  moment-sums  for  two  different  origins  equal  zero,  the  line  join- 
ing the  origins  not  to  be  parallel  to  the  forces.  Either  set  will  furnish  a 
solution  of  the  problem.    The  second  set  is  recommended,  and  the  origins 

of  moments  a  and  b  should  be  taken  on 

2000\b5.  loooibs.      Soooibs.  the  lines  of  action  of  the  two  unknown 

i 1      A       i B       forces.     For  example,  consider  the  beam 


R. 


:        /   >i<?'M-3->i<- 7' >         represented  in  Fig.  56  under  the  action 


^2    of  three  loads  (its  own  weight  neglected), 
Fig.  56  ^^'^  supported  at  A  and  B;  required,  the 

reactions  of  the  two  supports.  The  five 
forces  just  mentioned  constitute  a  system  in  equilibrium;  therefore,  taking 
moment  origins  on  Ri  and  R2  respectively,  and  assuming  that  Ri  and  R2  act 
upwards,  we  get 

SMi  =  2000  X  6  +  1000  X  2  —  3000  X  3  +  -??2  X  10  =  o, 
and         21/2  =  2000  X16  +  1000  X12  -f  3000  X  7  —  i?i  X  10  =  o. 


Art.  12 


45 


The  first  gives  Ri  =  —  $oo  pounds,  and  the  second  Ri  =  6500;  the  negative 
sign  means  that  R2  acts  downward  on  the  beam  and  not  upward,  as  as- 
sumed.    As  a  check  on  the  solution  we  try  whether  XF  =  o;  thus, 

—  2000  —  1000  —  3000  +  6500  —  500  =  o. 

The  graphical  solution  of  the  foregoing  problem  is  based  on  the  conditions 
that  the  force  and  the  string  polygon  for  the  forces  close;  the  process  of 
constructing  and  closing  the  polygons  determines  the  unknown  forces.  To 
illustrate  we  take  the  beam  shown  in  Figs.  56  and  57  and  determine  the 
reactions.  First,  the  force  polygon  should  be  drawn  as  far  as  possible, 
the  knowns  represented  first,  thus  AB,  BC,  and  CD  (Fig.  58)  representing 
the  2000,  the  1000,  and  3000  pound  forces  respectively;  then  the  lines  of 
action  should  be  lettered  to  correspond,  ah,  he,  and  cd  (Fig.  57).  If  R2,  say, 
is  taken  next,  it  would  be  lettered  DE,  and  Ri  would  be  EA,  since  the  force 


2000  Ibv.  1000  lbs.       ^000  lbs 


2000lb5-1000lbs.        ^OOO'"*- 


Fig.  S7 


Fig.  58 


Fig.  59 


polygon  for  all  must  close.  It  remains  now  to  locate  E;  this  can  be  done  by 
means  of  the  string  polygon.  (At  this  point  it  may  be  well  for  the  reader 
to  recall  the  significance  of  the  strings  of  a  string  polygon;  see  Art.  6.) 
The  polygon  may  be  started  at  any  point  on  any  of  the  lines  of  action  of 
the  forces  of  the  system;  if  it  be  started  at  i  (on  ah),  then  strings  oa  and 
oh  must  be  drawn  through  that  point;  oc  must  be  drawn  from  2  (where  oh 
cuts  he),  od  from  3  (where  oe  cuts  cd),  and  oe  from  4  (where  od  cuts  de)  and 
from  5  (where  oa  cuts  ea) ;  hence  the  closing  string  oe  passes  through  4  and 
5.  Finally,  the  ray  OE,  parallel  to  oe,  is  drawn,  thus  determining  E;  DE 
represents  R2,  and  EA  Ri.  Fig.  59  shows  another  solution;  Ri  is  taken  as 
the  fourth  force  DE\  and  R2  as  the  fifth  E'A. 

§  2.  We  take  this  opportunity  to  mention  a  class  of  problems  on  forces 
in  equilibrium,  not  parallel  necessarily,  which  cannot  be  solved  by  the 
principles  of  statics  alone,  and  are  therefore  called  statieally  indeterminate 
problems.  A  beam  resting  on  more  than  two  supports  furnishes  a  simple 
illustration;  thus,  let  it  be  required  to  determine  the  reactions  of  the  sup- 
ports {A,  B,  and  C)  on  the  beam  represented  in  Fig.  60,  due  to  the  two 


46  Chap,  n 

loads.  If  not  already  warned  of  the  difficulty  in  this  problem,  some  stu- 
dents would  probably  write  moment  equations  for  the  forces  in  equilibrium 
(Pi,  Pi,  Ri,  Ri,  and  Rz),  with  moment  origins  at  A,  B,  and  C,  and  then 
attempt   to  solve   the   equations  simultaneously   for   the   three    unknowns. 

Such  attempt  would  fail,  even  though  each 
jP.  |Pz  equation   would   be   correct,    because   the 

I '^ 1 ^ X-       three   would   not  be  independent  —  there 

IR,  'rj  'Rj     being  only  two  conditions  of  equilibrium 

Fig.  60  ^<^r  ^  system  of  the  kind  under  considera- 

tion (Art.  10  under  iv)  —  and  so  the  three 
equations  would  not  determine  the  three  unknowns.  Doubters  are  advised 
to  try  to  determine  Ri,  Ro,  and  R3  in  this  way  in  the  simple  case  where  the 
spans  and  the  loads  are  equal,  and  the  loads  are  applied  at  the  centers  of 
the  spans. 

How  may  one  determine  whether  a  given  problem  (a  force  system  in 
equilibrium  with  some  unknowns  required)  is  statically  determinate  or 
indeterminate?  A  complete  answer  to  the  question  is  beyond  the  scope 
of  this  book;  we  may  remark,  however,  that  statically  indeterminate  prob- 
lems commonly  arise  in  connection  with  structures  which  have  redundant 
or  superfluous  parts  or  supports,  by  which  is  meant  that  some  of  the  parts 
or  supports  are  not  strictly  necessary  for  the  equilibrium  of  the  structure. 
For  example,  in  Fig.  60  one  support  is  superfluous,  since  the  beam  on  tv/o 
supports  would,  if  strong  enough,  support  the  load.  No  statically  inde- 
terminate problems  are  given  in  this  book  without  notice;  but  the  student 
may  meet  a  force  system  in  equilibrium  containing  many  unknowns,  and 
he  is  now  reminded  that  it  is  futile  to  write  out  more  equilibrium  equations 
than  there  are  algebraic  conditions  of  equilibrium  for  the  system  under 
consideration  (Art.  10),  with  the  expectation  that  the  equations  if  solved  will 
determine  the  unknowns.  And  so  it  is  well  to  know  the  number  of  con- 
ditions of  equilibrium  for  each  class  of  force  systems. 

13.    Coplanar  Nonconcurrent  Nonparallel  Forces 

Principles  of  equilibrium  for  a  force  system  of  this  kind  are  developed  in 
Art.  10  under  (iv).  Their  use  will  be  explained  now  by  applying  them  to 
two  particular  common  problems. 

§  I.  Typical  Problem  (iv).  —  A  system  of  coplanar  nonconcurrent  non- 
parallel  forces  is  in  equilibrium,  and  all  except  two  are  wholly  known;  only 
the  line  of  action  of  one  of  these  two  and  a  point  in  that  of  the  other  are 
known,  and  it  is  required  that  these  two  be  determined  completely. 

The  algebraic  solution  of  this  problem  can  be  effected  by  means  of  any  one 
of  these  sets  of  equilibrium  equations: 

SF,  =  ZFy  =  XM  =  o;  2i?x  =  2M„  =  l^Mb  =  o;  2Ma  =  ZMt  =  Slfc  =  o. 


Art.  13 


47 


SO.OOOlbs 


Fig.  61 


For  an  example,  consider  the  roof  truss  represented  in  Fig.  61.     It  sustains 

two  loads,  3 5, 000  (weight  of  roof  and  truss)  and  50,000  pounds  (wind  pressure). 

The  left  end  of  the  truss  merely  rests  on  a 

wall,  but  the  right  end  is  fastened  to  a  wall; 

therefore  the  reaction  of  the  left-hand  wall 

must  be  vertical,  but  that  of  the  other  may 

be  inclined.    Let  it  be  required  to  determine 

these  reactions.    We  call  the  left  reaction  A , 

the  right  one  B,  and  the  inclination  of  B 

to  the  horizontal  6.     Then  the  first  set  of 

equilibrium  equations  gives  XMb  =  +35,000  X  45  +  50,000  X  (60  cos  30°)  — 

A  X  go  =  o,  or  A  =  46,400  pounds.     2F^  =  —Bco5d  +  50,000  sin  30°  =  o, 

and    2/^y  =  +B  sin  6  —  50,000  cos  30°  —  35,000  +  46,400  =  o;    these  solved 

simultaneously  give  B  =  40,500  pounds  and  0  =  51°  54'. 

For  algebraic  solutions,  it  is  generally  advisable  to  imagine  the  second 
unknown  force,  whose  point  of  application  is  known,  to  be  replaced  by  two 
(unknown)  components.  Then  the  problem  is  in  the  form  of  typical  problem 
(v)  (see  next  page).  Thus,  in  the  preceding  example  the  unknowns  would  be  A 
and,  instead  of  B  and  6,  B^  and  By.  After  finding  Bj,  and  By,  one  could  easily 
get  B  and  6. 

The  graphical  solution  of  this  problem  is  eftected  by  drawing  the  force  and 
the  string  polygons,  making  both  close  since  the  force  system  is  in  equilibrium. 

To  illustrate  we  use  the  preceding 
example.  We  first  draw  the  polygon 
ABC  (Fig.  62)  for  the  known  forces, 
and  continue  it  with  a  line  through 
C  parallel  to  the  left-hand  reaction. 
The  end  of  that  line,  as  yet  unknown, 
is  to  be  marked  D;  that  point  once 
determined,  then  DA  will  represent 
the  right-hand  reaction.  To  find  D  we  must  construct  a  string  polygon;  so 
we  next  mark  the  lines  of  action  of  the  several  forces  to  agree  with  the  nota- 
tion in  the  force  polygon,  choose  a  pole  0,  and  draw  the  rays  OA,  OB,  and 
OC.  To  make  use  of  the  known  point  i  of  the  fourth  force  (right-hand  reac- 
tion), the  string  polygon  must  be  begun  at  that  point.  The  string  oa  is  the 
one  to  draw  through  that  point  (to  ab),  and  then  ob  and  oc  as  shown.  The 
string  od  must  pass  through  points  i  and  4,  and  so  is  determined.  Next  we 
draw  the  ray  OD  (parallel  to  od),  and  thus  determine  D  (the  intersection  of 
CD  and  OD). 

The  following  special  graphical  method  is  simpler  in  principle  than  the  pre- 
ceding method:  Let  R  =  the  resultant  of  the  wholly  known  forces,  P  =  the 
force  whose  line  of  action  is  known,  and  Q  =  the  force  whose  point  of  applica- 
tion is  known.  Find  R,  and  then  imagine  the  wholly  known  forces  replaced 
by  R;  R,  P,  and  Q  would  be  in  equilibrium.     Now  a  balanced  three- force 


35,000  lb5. 


Fig.  62 


48 


Chap,  n 


50.0001b5.A( 


Fig.  63 


system  is  concurrent  or  parallel  (Art.  10,  §  2);  hence  if  R  intersects  P,  then 
Q  acts  through  that  point  of  intersection,  and  if  R  is  parallel  to  P,  then  Q  is 
also.  If  the  three  forces  are  concurrent,  then  determine  P  and  Q  from  the 
force  triangle  for  the  three  forces  as  explained  in  Art.  1 1 ;  if  they  are  parallel, 
determine  P  and  Q  as  explained  in  Art.  12.  To  illustrate,  we  use  the  data 
of  the  foregoing  example.  First  we  draw  AB  and  BC  (Fig.  63),  to  represent 
the  two  loads;  then  AC  represents  the  magnitude  and  direction  of  their 
resultant  R.     The  line  of  action  of  R  is  ac,  parallel  to  AC  and  passing  through 

the  intersection  of  ab  and  be.  (When 
the  wholly  known  forces  are  noncon- 
current  it  is  necessary  to  construct  a 
string  polygon  to  find  a  point  in  the  line 
of  action  of  R,  see  Art.  6.)  We  next 
extend  the  lines  of  action  of  R  and  P, 
and  join  their  intersection  with  the  point 
of  application  of  Q;  this  line  is  the  line 
of  action  of  Q.  Finally  we  complete  the  force  triangle  AC  DA  for  R,  P,  and 
Q;  then  CD  =  P  and  DA  =  Q. 

§2.  Typical  Problem  (v).  —  A  system  of  coplanar  nonconcurrent  non- 
parallel  forces  is  in  equilibrium,  and  all  the  forces  except  three  are  wholly 
known;  only  the  lines  of  action  of  these  three  are  known,  and  their  magni- 
tudes and  senses  are  required.* 

The  algebraic  solution  of  this  problem  can  be  effected  by  means  of  any  one 
of  these  three  sets  of  equilibrium  equations: 

SFx  =  SF„  =  2M  =  o;  2/^^  =  2Ma  =  Mlb  =  o;  or  l^Ma  =  ^Ah  =  ZMc  =  o. 

For  example,  consider  the  crane  represented  in  Fig.  64.  It  consists  of  a  post 
AB,  a.  boom  CD,  and  a  brace  EF;  the  post  rests  in  a  depression  in  the  floor 
below,  and  against  the  side  of  a  hole  in  the 
floor  above.  The  external  forces  acting  on 
the  crane  consist  of  the  load  W  (8  tons),  the 
weights  of  the  parts  named  (0.8,  0.9,  and  i.i 
tons  respectively),  and  the  reactions  of  the 
floors.  The  upper  floor  exerts  a  single  hori- 
zontal force  on  the  post;  the  lower  floor 
exerts  two  forces  on  the  post,  one  horizontal 
and  one  vertical.  Let  it  be  required  to  de- 
termine the  magnitudes  of  these  reactions. 
The  entire  external  system  of  forces  just 
described  is  in  equilibrium.  Calling  the  reactions  A,  B^,  and  By  respectively, 
then  the  first  set  of  equilibrium  equations  become:  2Ma  =  —  8  X  20  — 
0.9  X  II  -  I.I  X  7  +  5x  X  18  =  o,  or  B^  =  9.86;  XFj,  =  9.86  -  A  =  o,  or 
A  =  9.86;  XFy  =  By  —  8.0  —  0.8  —  0.9  —  I.I  =  o,  or  By  =  10.8  tons. 

*  If  the  three  unknown  forces  are  concurrent  or  parallel,  the  problem  is  indeterminate. 


'^^''mwi^ 


Fig.  64 


Art.  13 


49 


0.9tons 


etons 


Fig.  65 


The  general  graphical  solution  is  carried  out  as  follows:  Let  P,  Q,  and  S 
stand  for  the  three  forces  whose  lines  of  action  only  are  known.  Imagine  any 
two  of  these,  say  P  and  Q,  replaced  by  their  resultant  R';  one  point  in  that 
resultant  is  known,  the  intersection  of  P  and  Q.  Then  S,  R',  and  the  known 
forces  would  be  in  equilibrium,  and  the  given  problem  has  been  transformed 
to  typical  problem  iv.  So  we  first  determine  S  and  R',  as  explained  in  §  i, 
and  then  resolve  R'  into  two  components  parallel  to  P  and  Q;  these  compo- 
nents are  P  and  Q.  To  illustrate,  we  take  the  preceding  example,  and  we  call 
the  two  lower  reactions  P  and  Q,  and  the  upper  one  5  (Fig.  65).  The  re'sultant 
R'  of  P  and  Q  passes  through  the  lower 
end  of  the  post.  We  draw  the  polygon 
ABCDE  for  the  knowns,  and  continue 
it  with  a  line  parallel  to  S,  The  as 
yet  unknown  end  of  that  line  is  to  be 
marked  F;  that  point  once  determined, 
then  FA  will  represent  R' ,  since  the 
polygon  for  all  the  forces  must  close. 
To  find  F  we  must  construct  a  string 
polygon;  so  we  mark  the  lines  of  ac- 
tion of  the  several  forces  to  agree  with 
the  notation  in  the  force  polygon,  choose 
a  pole  0,  and  draw  rays  OA,  OB,  OC, 

OD,  and  OE.  The  string  polygon  must  be  begun  at  the  lower  end  of  the 
post,  the  point  of  application  of  FA  or  R'.  The  strings  to  pass  through 
that  point  are  of  and  oa  (Art.  6),  and  so  we  draw  oa  to  ab;  then  ob,  oc,  od,  and 
oe  as  shown.  Now  point  i  is  in  of,  and  point  6  is  also;  therefore  of  is  deter- 
mined. The  ray  OF  is  drawn  next  (parallel  to  of),  thus  determining  i^;  then 
EF  and  FA  represent  5  and  R',  as  already  stated.  Finally  we  draw  through 
F  a  vertical  and  through  A  a  horizontal;  then  FG  and  GA  represent  the 
vertical  and  horizontal  reactions  (P  and  Q)  of  the  lower  floor. 

The  following  special  graphical  method 
is  simpler  in  principle  than  the  preced- 
ing: First  we  determine  the  resultant  R 
of  the  wholly  known  forces;  R  and  the 
three  partly  unknown  forces  (P,  Q,  and 
S)  would  be  in  equilibrium.  The  special 
condition  of  equilibrium  for  four  such 
forces  is  that  the  resultant  R'  of  any 
pair  as  P  and  Q  balances  the  other  pair; 
hence  R'  and  the  other  pair  (R  and  S) 
are  in  equilibrium,  and  so  must  be  con- 
current or  parallel.  Next  we  solve  the 
system  R\  R,  and  S  (if  concurrent  by  Art.  11,  and  if  parallel  by  Art.  12). 
Finally  we  complete  the  force  polygon  for  R,  S,  P,  and  Q.     For  an  illus- 


FiG.  66 


50 


Chap,  n 


tration  we  take  the  preceding  example.  Let  the  two  lower  reactions  be 
called  P  and  Q,  and  the  upper  one  5  (Fig.  66).  The  resultant  R  of  the 
loads  is  I0.8  tons  acting  as  shown  (construction  for  R  is  indicated).  The 
resultant  R'  acts  through  point  i;  and,  since  R  and  .S  are  concurrent  at 
point  2,  R'  acts  through  point  2  also.  We  now  draw  the  force  triangle  AEFA 
for  R,  S,  and  R' ,  AE  representing  R;  then  EF  represents  S.  Finally  we  draw 
lines  from  A  and  F  parallel  to  Q  and  P,  thus  fixing  G\  and  then  FG  represents 
P,  and  GA  represents  Q. 

14.   Noncoplanar  Forces  in  Equilibrium 

§  I.  The  pirinciples  of  equilibrium  for  noncoplanar  forces  are  set  forth  in 
Art.  10  under  (v),  (vi),  and  (vii).  The  three  following  illustrations  deal  with 
concurrent,  parallel,  and  nonconcurrent  nonparallel  forces  respectively. 

(i)  A  heavy  body  W  (Fig.  67)  weighing  1000  pounds  is  suspended  from  a 
ring  over  the  center  of  a  street  60  feet  wide;    the  ring  is  supported  by  three 

ropes  OA,OB,  and  OC;  A  and  B  are  points 
on  the  face  of  a  building  as  shown,  and  C  is 
a  point  on  the  face  of  a  building  (not  shown) 
on  the  opposite  side  of  the  street,  OC  being 
perpendicular  to  the  face  of  the  buildings. 
Values  of  the  tensions  in  the  ropes  are  required. 
There  are  four  forces  acting  on  the  ring, — 
the  pull  of  1000  pounds,  and  the  pulls  of  the 
three  ropes  which  we  call  L,  M,  and  N  respec- 
tively; this  system  is  concurrent.  To  deter- 
mine the  unknown  forces  in  it,  we  use  the 
conditions  that  the  algebraic  sums  of  the  com- 
ponents along  three  rectangular  axes  equal 
zero;  as  axes  we  choose  a  vertical  line  and 
two  horizontal  lines,  one  parallel  and  one 
To  get  the  components  of  L,  M,  and  N,  we  need 
values  of  certain  angles:  A'OC'  =  tan-^  A'C'/OC'=  28°  4';  AOA'=  tan-i 
AA'/OA'=  30°  28';  B'OC^  tan-'  B'C'/0C'=s8°  40';  BOB'=  tan"'  BB'/OB' 
=  46°  11'.  The  X,  y,  and  z  components,  respectively,  of  L  are  L  cos  30°  28' 
sin  28°  4'=  0.405  L,  L  sin  30°  28'=  0.507  L,  and  L  cos  30°  28'  cos  28°  4'  = 
0.760  L;  of  M  they  are  il/ cos  46°  11'  sin  38°  40'=  0.4325  M,  M  sin  46°  ii'  = 
0.721  M,  and  M  cos  46°  11'  cos  38°  40'=  0.5405  M;  of  N  they  are  o,  o,  and 
N\  of  the  looo-pound  pull  they  are  o,  1000,  and  o.  The  algebraic  sums 
of  the  X,  y,  and  z  components  are 

—  0.405  L  -\-  0.4325  M  -\-  o  -\-  o  =0  , 
-{-0.507  L  -f  0.721  M  -1-  o  —  1000  =  o, 

—  0.760  L  —  0.5405  M  +  iV  -|-  o  =  o. 

Solving  these  equations  simultaneously,  we  find  that  L  =  846,  M=  792,  and 
N=  1072  pounds. 


Fig.  67 
transverse  to  the  street 


Art.  14 


SI 


Fig.  68 


(ii)  A  body  weighing  1000  pounds  is  suspended  from  the  ceiling  of  a  room 
by  means  of  three  vertical  ropes;  the  points  of  attachment  at  the  ceiling  lie 
at  the  vertices  of  an  equilateral  triangle  ABC  (Fig. 
68)  whose  sides  are  10  feet  long;  W  is  the  projection 
of  the  center  of  gravity  of  the  body  upon  the  ceiling. 
The  tension  in  each  rope  is  required.  We  call  the 
tensions  in  the  ropes  fastened  at  A,  B,  and  C,  respec- 
tively, L,  M,  and  N.  The  four  forces  acting  on  the 
body  constitute  a  parallel  system;  the  conditions  of 
equilibrium  for  such  are  that  the  sums  of  the  moments 
of  the  forces  about  any  three  coplanar  nonparallel  axes  perpendicular  to  the 
forces  equal  zero.  The  lines  AB,  BC,  and  CA  are  good  lines  to  choose  as 
axes  of  moments.  With  respect  to  these  lines  the  moment  equations  are 
respectively,  N  X  8.66  —  1000  X  2.10  =  o,  L  X  8.66  —  1000  X  4.15  =  o,  and 
M  X  8.66  —  1000  X  2.41  =0,  8.66  being  the  altitude  of  the  triangle.  Solu- 
tion of  these  equations  shows  that  L  =  479,  M  =  278,  and  N  =  243  pounds. 

(iii)  Fig.  69  shows  a  velocipede  crane.     The  crane  can  be  run  along  on  a 
single  rail  below,  tipping  being  prevented  by  two  overhead  rails  which  guide 

a  horizontal  wheel  mounted  on  the  top 
of  the  crane  post.  The  crane  weighs 
1.25  tons,  and  it  is  balanced  so  that  its 
center  of  gravity  is  in  the  axis  of  the 
post.  We  will  now  show  how  to  deter- 
mine the  supporting  forces  (exerted  by 
the  rails)  when  the  crane  supports  a 
load  of  1.5  tons 
and  the  jib  is 
swung  out  at 
right  angles  to 
the  rails  to- 
ward  the   left 

:  (Fig.  70). 

There  are 
three  support- 
ing forces  or 
reactions,  one 
on  each  wheel. 
Since  the  lower 
rail  is  level,  the 

crane  does  not  tend  to  roll,  and  there  is  no  reaction  of  the  rails  in  their 
direction.  The  reaction  of  the  upper  rail  is  directed  horizontally  and  evi- 
dently as  shown;  the  reaction  on  each  lower  wheel  has  two  components 
as  shown.     We  call  these  component  reactions  Ax,  Ay,  Bx,  and  5„,  and  the 


1^ 


Fig.  70 


'■Vi 


'v> 


upper  reaction  C.     The  external  system  of  forces  acting  on  the  entire  crane 


52 


Chap,  ii 


■-rV 


T-Y 


consist  of  the  reactions  named,  the  weight  of  the  crane,  and  the  load.  For 
noncoplanar  nonconcurrent  nonparallel  systems  there  are,  in  general,  six  con- 
ditions of  equilibrium,  but  this  system  has  only  five  because  there  are  no 
" z  forces  "  (see  the  figure).     The  five  conditions  of  equilibrium  are 

2F,  =A,  +  B,-C  =  o;  (i) 

2i^„   =+^„  +  ^„- 1.25- 1.5  =  0;  (2) 

2if^  =  ByX4-  AyX6  =  0;  (3) 

XMy  =  ^x  X  4  -  -^x  X  6  =  o;  (4) 

ZM,  =  C  X  16  -  1.5  X  6  =  o.  (5) 

From  (5)  it  follows  that  C  =  5.625  tons;  from  (i)  and  (4),  that  Bx  =  3.375 
tons,  and  Ax  =  2.25  ^tons;  from  (2)  and  (3),  that  By  =  1.65  tons,  and  Ay  = 
1. 10  tons. 

We  now  give  another  solution,  making  use  of  the  principle  that  if  the  forces 
of  a  system  in  equilibrium  be  represented  by  vectors,  then  the  projection  of 

the  vectors  on  any  plane  represents  a 
force  system  also  in  equilibrium  (see 
Art.  10  under  (vii)).  Fig.  71  shows 
such  projections  on  the  x-y,  y-z,  and  z-x 
planes  of  Fig.  70.  From  the  y-z  pro- 
jection (side  elevation),  2ilf a  =ByX 
10  —  2.75  X  6  =  o,  or  By  =  1.65  tons; 
and  2ifB  =  —  Ay  Xio  +  2.75  X  4  =  o, 
or  Ay  =  1. 10  tons.  From  the  x-y  projec- 
tion (end  elevation),  ZMa  =  C  X  16  — 
1.5  X  6  =  o,  or  C  =  5.625  tons.  From 
the  z-x  projection  (plan),  'ZMa  =  —  BxX 
^  10  -\-  5.625  X  6  =  o,  or  Bx  =  2.375  tons; 
and  27lfB  =  — ^i  X  10  +  5.625  X  4  =  o, 
or  Ax  =  2.25  tons. 

§  2.  A  noncoplanar  system  can  gen- 
erally be  solved  by  means  of  an  equivalent  coplanar  system.  This  indirect 
method  is  regarded  as  simpler  than  the  direct  one  when  the  forces  of  the  non- 
coplanar system  are  nonparallel.    The  two  following  examples  will  illustrate. 

For  one  example  we  use  the  data 
of  example  (i).  Instead  of  ropes  OA 
and  OB  (Fig.  67),  imagine  a  rope  00' 
in  the  plane  of  those  ropes,  and  also 
in  the  same  vertical  plane  with  COC. 
Such  a  rope  fastened  to  0  and  to  the 
building  at  0'  would  help  to  support 
the  ring  in  its  place,  8.nd  would  leave 


Plan. 


Fig.  71 


[--«■ -7H 


rB 


y 


/ 


■r^ 


/ 


V 


p./ 


C-e— 

N 


0,/ 


Il0001b5. 


A.- 


Ml 

T 

Fig.  72 


5-- 

\  '  / 


l\/ 


the  tension  in  OC  unchanged.     Thus  the  ring  would  be  acted  upon  by  three 
forceSj  — 1000,  iV,  and  the  pull  P  of  the  new  rope  (Fig.  72).     A  force  tri- 


Art.  14 


53 


angle,  FGHF,  for  these  forces  shows  that  the  pull  N  =  1070  and  P  =  1460 
pounds.  We  next  lay  out  the  ropes  OA,  OB,  and  00'  in  their  true  relations, 
and  then  we  resolve  the  pull  1460  in  the  imaginary  rope  into  components 
along  the  real  ropes.  Thus  we  lay  off  OQ  equal  to  1460,  and  then  on 
the  diagonal  OQ  complete  the  parallelogram  OMQN;  and  find  OM  and  ON, 
representing  the  tensions  in  the  real  ropes,  860  and  790  pounds. 

For  another  illustration  we  take  a  tripod  (Fig.  73),  shown  in  plan  and  eleva- 
tion. The  requirement  is  to  determine  the  forces  acting  at  the  top  of  each  leg 
of  the  tripod  due  to  a  load  of  1000  pounds.  On  account  of  this  load,  each  leg 
is  under  the  action  of  two  forces,  one  applied  at  each  end  of  that  leg,  and  so 
those  two  forces  act  along  the  axis  of  the  leg.  We  imagine  a  single  leg  in  the 
plane  of  any  two,  and  in  the  same  vertical  plane  with 
the  third,  to  replace  the  two;  thus  OD  to  replace  OA 
and  OB.  Then  there  would  be  three  forces  applied 
to  the  pin  at  0,  namely,  the  load  1000  pounds,  and 
the  supporting  forces  exerted  by  OC  and  OD.  So  we 
draw  a  force  triangle  for  these  three  forces  FGHF; 
it  shows  that  the  push  of  OC  is  GH  =  565,  and 
that  of  OD  is  HF  =650  pounds.  Next  we  lay  out 
the  other  pair  of  legs  and  the  imaginary  one  in  their 
true  relation  0"A",  0"B",  and  0"D",  and  make 
0"P  =  HF  =  650  pounds;  then  resolve  0"P  into 
two  components  along  the  pair  0"A"  and  0"B"  by 
means  of  a  parallelogram  0"MPN.  Thus  we  find  that 
0"M  and  0"N  represent  the  pushes  of  AO  and  BO, 
or  340  pounds. 


CHAPTER   III 

SIMPLE   STRUCTURES 

15.   Simple  Frameworks  (Truss  Type) 

§  I.  The  frames  herein  considered  consist  of  straight  members,  and  the 
axes  of  all  the  members  lie  in  one  plane;  such  are  called  plane  frames,  and 
the  plane  of  the  axes  is  called  the  plane  of  the  frame.  In  order  to  make  the 
axes  of  all  members  lie  in  one  plane,  and  the  truss  symmetrical  with  respect 
to  that  plane,  some  of  the  members  must  be  made  in  parts  or  with  forked 

ends.  For  example  see  Fig.  74,  which  shows  plan 
and  elevation  of  a  joint  of  a  frame  at  which  four 
members  are  pinned  together,  one  vertical  (double), 
one  diagonal  D  (single),  and  two  horizontals  Hi  and 
Hi  (each  double). 

Wooden  members  are  generally  bolted  together 
with  more  or  less  mortising;  steel  members  are  riv- 
eted together  or  joined  by  pins  through  holes  in  the 
members,  the  axes  of  pins  and  holes  being  perpen- 
dicular to  the  plane  of  the  frame.  All  frames  here 
considered  are  assumed  to  be  of  the  pin-connected 
type;  and,  furthermore,  it  is  assumed  that  each 
member  connects  only  two  joints,  that  is,  extends  from  one  joint  to  another 
but  not  also  to  a  third  one. 

In  such  pin-connected  frames,  the  lines  of  action  of  the  pin  pressures  (forces 
exerted  by  pins  on  the  members)  are  in  or  parallel  to  the  plane  of  the  frame. 
Thus,  the  resultant  pressure  of  the  pin  on  the  diagonal  member  D  (Fig.  74)  is 
clearly  in  the  plane;  the  pin  exerts  on  the  vertical  member  two  forces  which, 
on  account  of  the  symmetrical  arrangement,  are  equal,  parallel,  and  equally 
distant  from  the  plane,  and  therefore  the  resultant  of  these  two  forces  lies  in 
the  plane;  and  obviously  the  resultant  of  the  forces  exerted  by  a  pin  on  each 
horizontal  member  lies  in  the  plane.  Thus  all  resultant  pin  pressures  will  be 
regarded  as  lying  in  the  same  plane,  and  we  will  have  only  coplanar  forces  to 
deal  with  in  the  present  connection.  We  assume  that  the  pins  are  practically 
frictionless;  in  that  case  each  pin  pressure  acts  practically  normally  to  the 
surface  of  the  pin,  and  so  the  line  of  action  of  each  pressure  cuts  the  axis  of 
the  corresponding  pin. 

In  this  and  the  following  articles  we  assume  that  the  loads  are  applied  to 
the  frame  at  its  joints  only,  and  in  such  manner  that  the  line  of  action  of  each 
load  cuts  the  axis  of  the  pin  at  the  joint.     Then  each  member,  if  its  own  weight 

54 


Art.  is  55 

is  neglected,  is  subjected  to  forces  (pin  pressures  and  loads)  at  its  two  pin 
holes  only,  somewhat  as  shown  in  Fig.  75  or  Fig.  76,  where  P'  and  P"  denote 
pin  pressures  and  L'  and  L"  loads.  Let  R'  denote  the  resultant  of  P'  and  L\ 
and  R"  the  resultant  of  P"  and  L".     Since  R'  and  R"  balance,  each  acts  along 


— — —  p  ?'/  ;  NP" 


Tension  Compression 

R'  ^nn^  A        ^  ^,^|___  '^ "        R'  m  A       B  n  R" 

Fig.  75  Fig.  76 

the  axis  of  the  member,  and  hence  each  member  is  under  simple  tension  or 
compression.  Any  two  parts  of  the  member,  as  m  and  n,  exert  equal  and 
opposite  forces  upon  each  other;  A  (Figs.  75  and  76)  denotes  the  force 
exerted  on  m  by  n,  and  B  that  exerted  on  n  by  m.  Since  A  balances  R' ,  and 
B  balances  R" ,  A  and  B  also  act  along  the  axis  of  the  member.  And 
obviously,  if  R'  and  R"  are  pushes  (the  member  in  compression),  then  A 
and  B  are  pushes;  and  if  R'  and  R"  are  pulls  (member  is  in  tension),  then 
A  and  B  are  pulls.  And  conversely,  if  A  and  B  are  pushes,  then  the  member 
is  in  compression;  and  if  pulls,  then  in  tension.  By  stress  in  a  member  is 
meant  either  of  the  two  forces  which  two  portions,  as  m  and  n,  exert  upon 
each  other.*  We  are  now  ready  to  explain  a  method  for  determining  the 
stresses  in  the  members  of  a  simple  truss  due  to  given  loads ;  we  begin  with  an 
Example. — Fig.  77  represents  a  truss  supported  at  each  end;  the  angles 
equal  60  degrees;  it  sustains  two  loads  of  2000 

pounds  each  and  one  of  1000  pounds.     First,  '°°°1"'^-  ^^^^I'"*- 

it  is  necessary  to  ascertam  the  values  of  the  A    1         \    A 

reactions   A    and    B.      Since   all    the    external  /'"    \  /"'\ 

forces  acting  on  the  truss  (loads  and  reactions)        V  \,-.  /  \, . 

are  in   equilibrium,  ^Ma  ^  B  X  40  —  2000  X     A/  \         /  V/    \         /  \b 

A '  ' 1 i         '  I \ 

30  —  1000  X  10  —  2000  X  20  =  o,  or5  =  2750     T  ,^°°°|"^^-         ^       I 

pounds;    and    Svlfs  =  -^  X  40+         ^^' '^'^ ^^' "^ 

1000  X  30  +  2000  X  10  +  2000  X 

20  =  o,    or    A  =  2250.      IIF  =  2250  +  2750  —  2000  —  2000  — 

1000  =  o,  which  result  checks  the  computed  values  of  A  and  B. 

We  now  direct  our  attention  to  the  joint  A,  the  small  part  of 
truss  near  A  (or  "  pass  a  section  "  about  A  and  consider  that 
part  of  the  truss  within  the  section),  and  then  note  all  the  forces  acting  on 
that  part  (see  Fig.  78).     There  are  three  such  forces,  —  the  reaction  2250 

*  The  term  stress  is  defined  variously.  Some  writers  use  it  to  designate  the  forces  which 
any  two  different  bodies  or  any  two  parts  of  the  same  body  exert  upon  each  other;  that  is 
they  use  it  as  a  general  term  for  an  "action  and  reaction"  (Art.  11).  Most  engineers,  how- 
ever, use  the  term  in  a  more  restricted  sense  to  designate  the  force  which  one  part  of  a  body 
exerts  upon  an  adjacent  part  at  the  surface  of  division. 


56  Chap,  m 

pounds,  and  the  two  forces  exerted  upon  the  part  under  consideration  by  the 
remainder  of  the  truss;  they  are  marked  Fi  and  F2,  and  both  are  assumed  to 
be  pulls.*  This  part  of  the  truss,  as  well  as  every  other  part,  is  at  rest,  and 
so  the  three  forces  are  in  equilibrium.  Determination  of  the  unknown  forces 
Fi  and  Fi  presents  typical  problem  (i)  (Art,  11).  We  choose  the  algebraic 
method  for  solving:  'LFy  =  F2  sin  60°  +  2250  =  o,  or  F2  =  —2600;  the 
negative  sign  indicates  that  F2  is  really  a  push,  that  is,  the  stress  is  com- 
pressive. SFi  =  Fi  —  2600  cos  60°  =  o,  or  Fi  =  +1300;  the  positive  sign 
indicates  that  the  stress  is  tensile.  Passing  a  section  around  B,  and  consider- 
ing the  forces  acting  on  the  part  of  the  truss  within  the  section  (or  "  con- 
sidering forces  at  joint  5"),  we  get  Fig.  79.  The  forces  are  the  reaction  2750 
pounds  and  the  two  forces  exerted  on  the  part  under  consideration  by  the 
remainder  of  the  truss;  they  are  marked  F3  and  F4  and  are  assumed  to  b'e 
pulls.  Solution  of  this  three-force  system  shows  that  F3  =  +1588  (tension), 
and  F4  =—  3177  (compression). 

Next  we  might  discuss  joint  C,  D,  or  E  and  determine  two  more  stresses. 
Fig.  80  represents  joint  C  and  the  forces  acting  upon  it  so  far  as  known.    Stress 

_  ^  ^  1 1000  lbs  lOOOlbs.l 

\        \/  K    '    ^'=7\ 

I3p01b5.  Ny     1588^^5.  /         \  /         \ 

gyjOibsT  ZOOollba.  2600  lbs.    1444  lbs-  666  lbs.      31771bs. 

Fig.  79  Fig.  80  Fig.  81  Fig.  82 

in  CA  was  determined  to  be  a  tension  of  1300  pounds;  therefore  the  part  of 
CA  not  shown  in  the  figure  exerts  a  pull  of  1300  on  the  part  shown  as  indi- 
cated. Similarly,  the  part  of  CB  not  shown  in  the  figure  exerts  a  pull  of  1588 
on  the  part  shown  as  indicated;  F5  and  F^  are  assumed  to  be  pulls.  Solution 
of  this  five-force  system  shows  that  F5  =  +1444  (tension),  and  Fg  =  +866 
(tension).  Taking  joint  D  next,  we  get  Fig.  81,  four  forces  acting  on  the 
joint  (the  load,  and  the  three  forces  exerted  on  the  joint  by  the  remainder  of 
the  truss),  DA  was  found  to  be  under  a  compression  of  2600  pounds,  hence 
the  part  of  DA  not  shown  in  the  figure  acts  on  the  part  shown  as  indicated; 
CD  was  found  to  be  under  a  tension  of  1444  pounds,  hence  the  part  of  DC  not 
shown  in  the  figure  acts  on  the  part  shown  as  indicated;  F7  is  assumed  to  be 
a  pull.  ZFx  =  o  shows  that  F^  =  —2021  (compression);  and  writing  out 
2Fj,  we  find  that  it  equals  zero,  which  is  a  fair  check  on  the  computation. 
Fig.  82  represents  joint  E  and  all  the  forces  acting  upon  it,  as  already  deter- 
mined. If  2Fa;  =  o  and  SFj,  =  o  for  those  forces,  then  the  check  on  the  pre- 
ceding computations  is  satisfactory. 

*  In  simple  trusses  the  kind  of  stress  (tension  or  compression)  in  any  member  is  apparent. 
When  the  kind  is  not  apparent,  we  might  follow  the  suggestion  in  the  footnote,  page  41. 
But  for  uniformity  we  will  always  assume  the  force  to  be  a  pull.  Then,  according  to  the 
footnote,  the  force  is  actually  a  pull  or  a  push  (and  the  stress  is  tensile  or  compressive),  ac- 
cording as  its  computed  value  is  positive  or  negative. 


27501bil 


AxT.  IS 


57 


5000  I  lbs. 


5000 1  lbs. 


Directions. — The  foregoing  method  for  "analyzing  a  truss"  (determin- 
ing the  stresses  in  its  members)  can  be  formulated  into  brief  directions 
as  follows:  (i)  Determine  the  reactions  (supporting  forces)  on  the  truss  if 
possible.  (2)  Consider  a  joint  at  which  there  are  only  two  unknown  forces, 
and  then  determine  those  two.  (3)  Repeat  (2)  again  and  again  until  all 
stresses  have  been  determined.  (These  directions  do  not  pro\ide  for  a 
certain  contingency  which  may  arise;  see  §  2  for  a  case  and  directions  for 
meeting  it.) 

We  now  give  illustration  of  truss  analysis  by  this  method  but  omitting  the 
computations;   they  should  be  supplied  by  the  student.     The  truss  shown  in 
Fig.  83  will  be  used;  it  is  supported  at  each  end,  and  supports  three  loads  of 
5000  pounds  as  shown.      Obviously  each 
reaction    equals    one-half    the    total    load. 
On  joint  A  there  are  three  forces  (the  re- 
action, and  the  stresses  in  AD  and  AE)\ 
solving  that  force  system  we  find  that  the 
first  stress  =  15,000  pounds   compression, 
and  the  second  =  13,000  tension.    On  joint 
D   there   are   four   forces  (the  load   5000  ^^'    ^ 

pounds,  the  stress  \n  AD  =^  15,000  pounds,  and  the  stresses  in  DE  and  DC 
unknown);  solving  that  system,  we  find  that  the  stress  in  DE  =  4335  pounds 
compression,  and  that  in  DC  =  12,500  compression.  On  joint  E  there  are 
four  forces  (the  stress  in  AE  =  13,000  pounds,  the  stress  in  DE  =  4335 
pounds,  and  the  stresses  in  EC  and  EG  unknown) ;  solving  the  system,  we 
find  that  the  stress  in  EC  =  4335  pounds  tension,  and  that  in  EG  =  8667 
tension. 

§  2.  We  now  explain  the  contingency  or  diflSculty  mentioned  in  the  fore- 
going directions  and  how  to  meet  it;   the  truss  shown  in  Fig.  84  furnishes  an 

illustration.  Following  the  directions, 
we  determine  the  reactions  Ri  and  R2, 
2800  pounds  and  2400.  Then  we  take 
joint  A,  and  find  stresses  in  AB  and 
-43"  to  be  3960  (compression)  and  2800 
(tension)  respectively;  next  we  take 
joint  G,  and  find  stresses  in  GF  and 
GI  to  be  3400  (compression)  and  2400 
(tension)  respectively.  No  joint  re- 
mains  at  which    there   are   only  two 


600 1  lbs 


800,  lbs 


SOOllbb. 


aooiibs. 


->i<- /e'--->l<-— - 

Fig.  84 

unknown  stresses,  and  the  difiiculty  is  already  met.  Now  if  in  some  way  we 
could  ascertain  the  stress  in  almost  any  other  member,  then  we  could  con- 
tinue to  apply  the  rule.  For  example,  if  we  knew  the  stress  in  HB,  HJ,  or 
HI,  then  consideration  of  joint  H  would  determine  the  two  unknown  stresses 
there;  consideration  of  joint  B  would  give  stresses  in  BJ  and  BC;  considera- 
tion of  joint  C  would  give  stresses  in  CJ  and  CD,  etc.     Now  there  is  a  way  to 


-o  Chap,  in 

ascertain  the  stresses  in  CD,  JD,  and  HI,  — hy  passing  a  section  through 

those  members,  and  solving  the  force  system  acting  upon  either  portion  of 

the  truss.     Fig,  85  represents  the  left-hand  portion  and  all  the  forces  acting 

^     upon  it;   namely,  the  three  loads,   the  left  reaction, 

,    ^7     and  the  forces  which  the  right-hand  part  exerts  (^i,  ^2, 

^°°'"'    /ss  and  53,  assumed  to  be  pulls).     Solution  of  this  force 

system  presents  typical  problem  (v)   (Art.   13).     To 

determine  Si,  for  example,  we  take  moments  about 

the  intersection  of  ^2  and  S3  (or  joint  D),  and  find 

^   Si=  1600  pounds  tension.     Then  having  determined 

z&oolibs.        1200 jibs.        s^  we  proceed  as  in  the  foregoing  examples. 

Fig.  85  In  order  to  determine  the  stress  in  any  particular 

member  of  a  truss  the  following  direction  may  be  tried:  Imagine  the  truss 
separated  into  two  distinct  parts  ("  pass  a  section  "  through  the  truss);  pass 
it  in  such  a  way  that  the  member  under  consideration  is  one  of  the  members 
cut  by  the  section,  and  so  that  the  system  of  forces  acting  on  one  of  the  two 
parts  is  solvable  for  the  desired  stress;  then  solve  the  system  for  the  desired 
stress.  (The  system  of  forces  acting  on  one  part  of  the  truss  consists  of  the 
loads  and  reactions  on  that  part,  and  the  forces,  or  stresses,  which  the  other 
part  exerts  upon  it.  In  plane  trusses  this  system  is  always  coplanar;  it  can  be 
solved  if  it  is  concurrent  with  not  more  than  two  unknowns,  or  if  it  is  non- 
concurrent  with  not  more  than .  three  unknowns,  provided  that  the  three 
unknowns  are  not  parallel  nor  concurrent.) 

Foregoing  direction  may  be  applied  not  only  to  bridge  over  the  difiiculty 
sometimes  met  in  connection  with  directions  in  §  i,  but  also  when  it  is  desired 
to  determine  the  stress  in  a  particular  member  quite  directly  without  first 
computing  stresses  in  several  other  members.  For  example,  let  it  be  required 
to  determine  the  stress  in  BC  (Fig.  86),  the  truss  being  supported  at  its  ends, 
span  ^£  =  32  feet,  rise  CG"  =  8  feet,  and  five  loads  as  shown.     Obviously 


Fig.  86 


1000  lbs. 
Fig.  87 


each  reaction  equals  4000  pounds.  A  section  cutting  BC,  BG,  and  OF  gives 
a  left-hand  part  of  the  truss  with  its  external  forces  as  shown  in  Fig.  87.  The 
force  system  can  be  solved  for  the  desired  stress;  taking  moments  about  the 
intersection  of  ^2  and  ^3  (joint  G),  we  get  -  5i  X  8  X  cos  26°  34'  —  4000  X 
16  -f  3000  X  8  =  o,  or  5i  =  —  5600,  the  negative  sign  indicating  that  Si  is 
compressive  and  not  tensile,  as  assumed  in  the  moment  equation. 


Art.  1 6 


59 


§  3.  Warning  is  here  given  that  not  all  trusses  can  be  analyzed  by  the 
principles  of  statics  alone,  as  in  the  preceding;  that  is  to  say,  there  are 
trusses  that  are  statically  indeterminate.  Only  the  so-called  complete  or  per- 
fect trusses  are  always  statically  determinate;  beside  these  there  are  incom- 
plete trusses,  and  trusses  with  redundant  members. 

A  pin-connected  triangle  (Fig.  88)  is  the  simplest  complete  truss;  it  is 
indeformable  and  has  no  superfluous  or  redundant  members.  Adding  two 
more  members  makes  a  complete  truss  of 
two  triangles;  and  each  addition  of  two 
members  as  shown  extends  the  truss  and 
leaves  it  complete.  If  m  =  number  of 
members,  and  j  =  number  of  joints,  then 
for  a  complete  truss,  m  =  2  j  —  t,.  A 
pin-connected  quadrilateral  (Fig.  89)  is 
the  simplest  incomplete  truss;  it  is  deformable  and  requires  the  addition  of 
one  or  more  members  to  make  it  complete.  For  an  incomplete  truss, 
w  <  27  —  3.  A  pin-connected  quadrilateral  with  two  diagonal  members 
(Fig.  90)  is  the  simplest  truss  with  a  superfluous  or  redundant  member;  it 
is  indeformable  and  would  be  so  with  any  member  removed.  For  a  truss 
with  a  redundant  member  m  >  2  j  —  t,-  Figs.  91,  92,  and  93  are  other 
examples  of  the  three  classes  of  trusses  described. 


Fig.  88 


Fig.  89  Fig.  90 


Fig.  qi 


Fig.  92 


Fig.  93 


In  the  foregoing  it  is  assumed  that  the  trusses  are  pin-connected,  and  that 
each  member  can  sustain  tension  or  compression  as  called  upon  by  the  loading. 
For  a  classification  not  so  restricted  as  this  one,  readers  are  referred  to  stand- 
ard works  on  Structures.* 


16.   Graphical  Analysis  of  Trusses;  Stress  Diagrams 

§  I.  Graphical  methods  are  especially  well  adapted  for  analyzing  trusses. 
As  in  the  algebraic  methods  of  the  preceding  article,  we  imagine  the  truss 
separated  into  two  parts,  and  direct  our  attention  to  the  external  forces  acting 
upon  either  part.  Graphical  instead  of  algebraical  conditions  of  equilibrium 
are  then  applied  to  these  forces  to  determine  the  unknowns.  The  notation 
for  graphical  work  described  in  Art.  2  can  be  advantageously  systemized  as 
follows:  Each  triangular  space  in  the  truss  diagram  is  marked  by  a  lower- 
case letter,  also  the  space  between  consecutive  lines  of  action  of  the  loads  and 
reactions  (Fig.  94) ;   then  the  two  letters  on  opposite  sides  of  any  line  serve  to 

*  Johnson,  Bryan,  and  Turncaure's  Modern  Framed  Structures. 


6o 


Chap,  hi 


lOOOMbs. 


1000  lbs. 


designate  that  line,  and  the  same  capital  letters  are  used  to  designate  the 
magnitude  of  the  corresponding  force.     This  scheme  of  notation  is  a  great 

help  in  graphical  analyses  of  trusses. 

As  an  illustration  we  determine  the 
stress  in  each  member  of  the  truss  of 
Fig.  94.  Evidently  each  reaction 
equals  one-half  the  load,  or  2000 
pounds.  We  "  pass  section  "  a,  and 
consider  the  forces  acting  on  the  left- 
hand  part  of  the  truss  (Fig.  95) ;  they  are  the  load  500  pounds,  the  reaction  2000 
pounds,  and  the  stresses  cd  and  da.  Since  those  forces  are  in  equilibrium,  their 
polygon  closes;  in  constructing  it,  the  unknowns  will  be  determined.  Beginning 
with  the  knowns,  yl.B  is  drawn  to  represent  2000  pounds,  BC  to  represent 
500  pounds ;  and  then  a  line  from  A  (or  C)  parallel  to  the  line  of  action  of  one 
unknown,  and  a  line  from  C  (or  A)  parallel  to  the  other,  are  drawn.  The  last 
two  lines  determine  D  (or  D'),  and  the  closed  polygon  is  A  BCD  A  (or  A  BCD' A) ; 
hence  the  forces  in  the  members  cd  and  ad  are  represented  by  CD  and  DA 
(3000  and  2600  pounds)  respectively.  It  is  seen  from  the  force  polygon  that 
CD  is  a  push,  and  DA  is  a  pull;  hence  the  members  cd  and  ad  are  in  com- 
pression and  tension  respectively. 


500  lbs 


b|a 
EOOOlbs 


D-- 


Fig.  95 


B 


->--'A 


3000  lbs. 


Fig.  96 


We  may  next  pass  section  (3,  and  consider  the  forces  acting  on  the  smaller 
(and  simpler)  part  of  the  truss  (Fig.  96);  they  are  the  load  1000  pounds,  the 
stress  3000  pounds  (compressive),  and  the  stresses  fe  and  de.  Their  force 
polygon  may  be  drawn  thus:  DC  to  represent  3000  pounds  (compression), 
CF  to  represent  1000  pounds,  a  line  from  F  parallel  to  one  of  the  unknowns, 
and  one  from  D  parallel  to  the  other.  The  last  two  lines  determine  E,  and 
the  force  polygon  is  DCFED;  hence  the  forces  in  the  members  fe  and  ed 
are  represented  by  FE  and  ED  (2500  and  866  pounds) 
respectively.     Both  members  are  in  compression. 

We  next  pass  section  7,  and  consider  the  forces  acting 
on  the  smaller  part  of  the  truss  (Fig.  97) ;  they  consist 
of  the  stress  2600  pounds  (tension),  the  stress  866  pounds 
(compression),  and  the  stresses  eg  and  ga.  Their  force 
polygon  may  be  drawn  thus:  AD  to  represent  2600  pounds 
(tension),  DE  to  represent  866  pounds  (compression), 
a  line  from  E  parallel  to  one  of  the  unknowns,  and  a  line  from  A  parallel  to 
the  other.    The  last  two  lines  determine  G,  and  the  force  polygon  is  A  DEC  A; 


866  lbs. 


^/a 


26001b5.d 
^        a 


^L 


.2— 
a 


D 


\? 


E 
Fig.  97 


AsT.  id  6l 

hence  the  forces  in  the  members  eg  and  ag  are  represented  by  EG  and  GA 
(866  and  1732  pounds)  respectively.  Each  member  is  in  tension.  On  account 
of  the  symmetry  of  the  truss  and  loading,  the  forces  in  the  remaining  mem- 
bers are  now  known. 

In  drawing  the  force  polygon  for  all  the  external  forces  on  the  part  of  a 
truss  included  within  a  section  about  a  joint,  it  will  be  advantageous  to  repre- 
sent the  forces  in  the  order  in  which  they  occur  about  the  joint.  A  force 
polygon  so  drawn  will  be  called  a  polygon  for  the  joint;  and  for  brevity,  if 
the  order  taken  is  clockwise  the  polygon  will  be  called  a  clockwise  polygon, 
and  if  counterclockwise  it  will  be  called  a  counterclockwise  polygon.  ABC  DA 
(Fig.  95)  is  a  clockwise  polygon  for  joint  b  of  Fig.  94;  A  BCD' A  is  a  force 
polygon  for  the  "forces  at  joint  i,"  but  it  is  not  a  polygon  for  the  joint,  be- 
cause the  forces  are  not  represented  in  the  polygon  in  the  order  in  which  the 
forces  occur  about  the  joint.  The  student  should  draw  the  counterclockwise 
polygon  for  the  joint,  and  compare  with  ABCDA. 

If  the  polygons  for  all  the  joints  of  a  truss  are  drawn  separately  as  in  the 
preceding  illustration,  then  the  stress  in  each  member  will  have  been  repre- 
sented twice.  It  is  possible  to  combine  the  polygons  so  that 
it  will  not  be  necessary  to  represent  the  stress  in  any  mem- 
ber more  than  once,  thus  reducing  the  number  of  lines  to 
be  drawn.  Such  a  combination  of  force  polygons  is  called 
a  stress  diagram.  Fig.  98  is  a  stress  diagram  for  the  truss 
of  Fig.  94  loaded  as  there  shown.  Comparing  the  part  of 
the  stress  diagram  consisting  of  solid  lines  with  Figs.  95, 
96,  and  97,  it  is  seen  to  be  a  combination  of  the  latter  three 
figures.  It  will  also  be  observed  that  the  polygons  are  all 
clockwise  polygons;  counterclockwise  polygons  also  could  be  combined  into 
a  stress  diagram. 

Directions  for  constructing  a  stress  diagram  for  a  truss  under  given  loads: 

(i)  Letter  the  truss  diagram  as  already  explained. 

(2)  Determine  the  reactions.  (In  some  exceptional  cases  this  stage  may  or 
must  be  omitted;  also  stage  (3).     See  §  2  for  two  illustrations.) 

(3)  Construct  a  force  polygon  for  all  the  external  forces  applied  to  the  truss 
(loads  and  reactions),  representing  them  in  the  order  in  which  their  points  of 
application  occur  about  the  truss,  clockwise  or  counterclockwise.  (The  part 
of  that  polygon  representing  the  loads  is  called  a  load  line.) 

(4)  On  the  sides  of  that  polygon  construct  the  polygons  for  all  the  joints. 
They  must  be  clockwise  or  counterclockwise  ones,  according  as  the  polygon 
for  the  loads  and  reactions  was  drawn  clockwise  or  counterclockwise.  The 
first  polygon  drawn  must  be  for  a  joint  at  which  but  two  members  are  fastened; 
the  joints  at  the  supports  are  usually  such.  Next  the  polygon  is  drawn  for  a 
point  at  which  not  more  than  two  stresses  are  unknown;  that  is,  of  all  the 
members  fastened  at  that  joint  the  forces  in  not  more  than  two  are  unknown. 
Then  the  next  joint  at  which  not  more  than  two  stresses  are  unknown  is  con- 


62 


Chap,  iii 


sidered,  etc.,  etc.     (These  directions  do  not  provide  for  a  certain  diflSculty 
which  may  arise;  see  §  2  for  a  case  and  directions  for  handling  it.) 

To  illustrate  the  foregoing  directions  we  analyze  the  truss  represented  in 
Fig.  99;    it  sustains  four  loads  (600,  1000,  1200,  and  1800  pounds),  and  is 


(000 


1800 


:s 


H 


supported  at  its  ends.  Supposing  the  reactions  to  have  been  determined,  we 
draw  the  force  polygon  for  the  loads  and  reactions  ABC  DBF  A,  at  the  left;  it 
is  a  clockwise  polygon.  We  may  begin  by  drawing  the  clockwise  polygon  for 
joint  I  or  2;  for  the  former  it  is  FABGF*  Member  hg  is  therefore  in  com- 
pression and  gf  in  tension.  Next  we  may  draw  the  clockwise  polygon  for 
joint  2,  3,  or  4;  for  the  joint  2  it  is  CDEHC.  Member  ch  is  in  compression 
and  eh  in  tension.  For  joint  3,  the  polygon  is  HEFGH,  and  member  gh  is  in 
tension.  If  the  work  has  been  correctly  and  accurately  done,  the  line  GH  is 
parallel  to  gh. 

§  2.  There  are  exceptional  cases  not  covered  by  the  foregoing  directions. 
In  case  the  reactions  cannot  be  determined  in  advance,  the  stress  diagram  can 

still  be  drawn  if  the  truss  is  statically  determinate.     Fig. 
ii  100  represents  such  a  case,  the  truss  being  pinned  to  its 

supports.     The  diagram  can  be  constructed  by  drawing  in 

succession  the  proper  polygons  (all  clockwise  or  counter- 
|r    clockwise)  for  joints  i,  2,  3,  and  4.     Then,  if  desired,  the 

reactions  can  be  determined  by  drawing  the  polygons  for 

joints  5  and  6. 
Fig.  loi  represents  a  case  where  the  reactions  can  be 
determined  at  stage  (2)  of  the  analysis,  but  determina- 
tion of  the  reactions  is  not  essential  for  the  construction 
of  the  stress  diagram.  The  truss  is  supported  by  a  shelf 
A  and  a  tie  B.     The  stress  diagram  can  be  constructed  Fig.  ioi 

by  drawing  in  succession  proper  polygons  for  joints  i,  2,  3,  4,  and  5.  The 
reaction  at  B  is  determined  by  the  polygon  for  joint  5;  that  at  A  by  the 
polygon  for  joint  6. 

*  The  student  is  urged  to  make  sketches  of  the  bodies  (parts  of  truss)  upon  which  the 
forces,  whose  polygons  are  being  drawn,  act.  A  force  acting  upon  the  "cut"  end  of  a  mem- 
ber and  toward  the  joint  is  a  pu'ih,  and  the  stress  in  the  member  is  compressive;  if  the  force 
acts  away  from  the  joint,  it  is  a  pull,  and  the  stress  is  tensile. 


Fig.  100 


Art.  i6  ^3 

Fig.  1 02  shows  a  truss  the  analysis  of  which  is  not  fully  provided  for  in  the 
directions.  Thus,  suppose  that  the  reactions  have  been  determined;  the 
polygon  for  joint  i  may  be  drawn  first,  next  that  for  joint  2,  and  then  that  for 
joint  3.  Similarly  the  polygons  for  joints  i',  2',  and  3'  can  be  drawn;  but 
then  no  joint  remains  at  which  there  are  but  two  unknown  stresses,  and  so 
no  more  polygons  can  be  drawn,  as  yet.  If  in  any  way  the  number  of  un- 
known stresses  at  a  remaining  joint  could  be  reduced  to  two,  then  the  polygon 
for  that  joint  could  be  drawn,  and  the  stress  diagram  could  be  completed. 
Thus,  if  the  stress  in  ij,  jm,  or  mf  could  be  determined,  then  the  polygon  for 
joint  4  could  be  drawn,  and  then  those  for  5,  6,  7,  and  8. 

1000  I  lbs. 

1000  Ibi.  1000  lbs 

I-  ig     . 

1000 lbs.     I  \  VA\^  1000 lbs. 


Fig.  102 

The  difficulty  here  pointed  out  is  just  like  that  mentioned  under  the  direc- 
tions in  §  I  of  the  preceding  article.  It  may  be  met  by  means  of  the  direction 
in  §  2  of  that  article,  which  explains  how  to  determine  the  stress  in  a  par- 
ticular member  quite  directly  and  independently  of  any  stress  diagram  or 
polygons  for  joints.  Thus  to  determine  the  stress  in  w/we  pass  a  section  as  a, 
and  solve  the  external  system  of  forces  (including  stresses  in  the  members 
cut)  which  acts  upon  either  part  of  the  truss  for  the  desired  stress.  Then  we 
proceed  with  the  stress  diagram  as  already  pointed  out.  There  are  other 
ways  of  meeting  the  difficulty  presented  in  this  form  of  truss,  but  that  here 
explained  is  quite  general  and  can  be  applied  readily  to  other  forms. 

We  will  now  explain  this  matter  in  detail,  using  the  same  truss.  Evidently 
each  reaction  equals  one-half  the  total  load.  ABCDEE'D'C'B'A  'FA  is  a  clock- 
wise polygon  for  the  loads  and  reactions.  The  polygon  for  joint  i  is  FABGF; 
that  for  joint  2  is  GBCHG;  that  for  joint  3  is  FGHIF.  The  polygons  for 
joints  i',  2',  and  3'  are  B'A'FG'B',  C'B'G'H'C,  and  H'G'FI'H'  respectively. 
The  forces  acting  on  the  part  of  the  truss  to  the  left  of  section  a  are  the  loads 
at  joints  i,  2,  5,  and  6,  the  left  reaction,  and  the  forces  exerted  on  the  left  part 
of  the  truss  by  the  right  (stresses  el,  Im,  and  mf).  This  system  may  be  solved 
graphically  or  algebraically;  the  algebraic  method  is  much  the  simpler,  arms 
of  forces  being  scaled  from  the  truss  drawing.  Thus  to  ascertain  the  stress 
mf,  we  take  moments  about  the  intersection  of  el  and  Im,  and  get  1000  X 


64 


Chap,  rn 


7.5  +  1000  X  15  +  1000  X  22.5  +  500  X  30  —  4000  X  30  —  {mf)  X  17.5  =  o, 
or  mf  =  3425  (tension).  Next  we  represent  the  stress  mf  in  its  proper  place 
in  the  stress  diagram  at  MF,  and  then  draw  the  polygon  for  joint  4;  it  is 
MFIJM.     Completion  presents  no  difl&culties. 

17.   Simple  Frameworks  (Crane  Type) 

The  frames  here  considered,  like  the  trusses  of  the  preceding  articles,  are 
plane  and  symmetrical  with  respect  to  the  plane  of  the  frame.  For  example, 
the  crane  represented  in  Fig.  103  consists  of  a  post  MN,  a  boom  PQ,  and  a 
brace  KQ;  the  boom  consists  of  two  pieces  between  which  the  post  and  the 
brace  lie,  and  the  brace  is  forked  at  its  lower  end  by  means  of  side  pieces 
and  straddles  the  post.  Like  the  trusses,  these  frames  are  assumed  to  be  pin 
connected,  the  pins  being  practically  frictionless.  Thus  each  pin  pressure 
lies  in  the  plane  of  the  frame,  and  the  line  of  action  cuts  the  axis  of  the  pin. 

Unlike  the  trusses,  these  frames  may  include  a  member  which  is  pinned  to 
others  at  more  than  two  points;  the  loads  also  on  these  frames  are  applied 
anywhere,  not  at  the  joints  necessarily.  The  result  of  these  conditions  is 
that  the  stress  in  any  member  of  the  frame  is  generally  not  a  simple  tension 
or  compression,  the  member  being  bent  as  well  as  stretched  or  shortened. 
We  will  not  attempt  to  determine  the  stresses  in  the  members  of  these  frames 
but  limit  the  discussions  to  a  determination  of  the  forces  which  act  upon 
each  member,  the  pin  pressures,  reactions  of  supports,  etc. 

In  general  the  pressure  of  a  pin  on  a  member  does  not  act  along  the  axis 
of  that  member.    Take,  for  example,  the  brace  (diagonal)  (Fig,  103);   it  is 


,    N 


Fig.  103 


acted  upon  by  three  forces,  —  its  own  weight  W  and  the  pin  pressures  K  and  Q. 
These  three  forces  must  be  concurrent  or  parallel  (Art.  10,  §  2).  If  they  are 
concurrent,  then  neither  K  nor  Q  is  axial  or  else  both  are;  but  obviously  both 
K  and  Q  cannot  be  axial  and  then  balance  W ,  and  so  neither  acts  axiaUy.  If 
they  are  parallel,  then  neither  K  nor  Q  acts  axially. 

In  some  consideration  of  frameworks,  the  weights  of  some  or  all  members 
are  negligible  in  comparison  with  other  forces  (loads)  which  act  upon  the 
frame,  and  so  we  may  have  to  do  with  a.  member  acted  upon  by  only  two 
forces,  —  pin  pressures.    On  such  a  member,  the  pin  pressures  do  act  along  the 


Art.  17 


65 


axis  of  that  member,  since  the  pressures  balance  each  other  and  so  must  be 
colinear  (Fig.  103). 

"  Analysis  of  a  crane  "  means  the  determination  of  every  force  (magnitude 
and  direction)  acting  on  each  part  or  member  due  to  weight  of  the  crane  or 
loads  on  it  or  both.  The  general  method  of  procedure  may  be  briefly  summa- 
rized as  follows:  (i)  Make  a  sketch  of  the  entire  crane,  and  represent  as  far 
as  possible  all  the  external  forces  acting  upon  it;  apply  the  appropriate  con- 
ditions of  equilibrium  to  the  force  system,  and  then  determine  as  many  of 
the  unknowns  as  possible.  (2)  Make  a  sketch  of  a  member  or  of  a  combina- 
tion as  they  are  on  the  crane,  and  represent  as  far  as  possible  all  the  external 
forces  acting  on  it;  then  apply  the  appropriate  conditions  of  eqmUbrium  to 
the  force  system,  and  then  determine  as  many  of  the  unknowns  as  possible. 
(3)  If  other  forces  remain  to  be  determined,  then  continue  as  directed  in 
(2),  bearing  in  mind  the  law  of  "  action  and  reaction  "  (Art.  11).  We  will 
now  give  two  examples  of  analysis  employing  both  algebraic  and  graphic 
methods. 

Example  (i).— We  analyze  the  crane  represented  in  Fig.  103;  the  crane 
is  supported  at  M  and  N  by  sockets  in  the  ceihng  and  floor.  MN  =18, 
PQ  =  14,  MP  =  NK  =  3  feet;  it  bears  a  load  of  8  tons  on  the  boom  at 
16  feet  from  the  axis  of  the  post;  weights  of  members  neglected.     Fig.  104 


rA— 
to 

^>r 

OTO 

p 

/i" 

t\j 

^  — .- 

.,4' 

/...-M 

1 

/^ 

* 
» 

/ 

t 


Py 


/ 


7.  U  tons 
P 


8  tons 


K 


V 


K 

7.lltons 


■©"z' 


n!x 


.0- 


^<^'' 


Nx|Nj 


N 
8  I  tons 


:t; 


Fig.  104 


(at  the  left)  represents  the  entire  crane  with  all  external  forces,  the  senses  of 
the  reactions  being  quite  obvious.  The  solution  of  this  system  falls  under 
Art.  13.  Silfiv  =  o  gives  M  =  7.1 1  tons;  since  'ZF^  =  o,  A'^  =  7.1 1  tons; 
and  since  SFy  =  0,  Aj,  =  8  tons.  We  sketch  the  brace  KQ  next.  Since 
it  is  a  two-force  member,  the  pin  pressures  K  and  Q  are  axial,  equal,  and 
obviously  have  senses  as  shown.  The  common  value  of  K  and  Q  cannot  be 
determined  from  a  consideration  of  their  equihbrium.  Next  we  sketch  the 
boom.  Q  on  the  boom  and  Q  on  the  brace  constitute  an  action  and  reaction, 
and  so  are  coHnear,  opposite,  and  equal;  the  pressure  at  P  is  unknown  in 
direction,  and  in  an  algebraic  solution  can  be  dealt  with  most  easily  through 
its  components  Px  and  Pj,,  senses  guessed  at.     Solution  of  this  system  falls 


66 


Chap,  hi 


under  Art.  13.  Since  'LMp=  o,  Q  =  14.05  tons;  since  2F;,  =  o,  P^  =  10.67 
tons;  since  XFy  =  o,  Py=  -1.14  tons,  the  negative  sign  indicating  that  Py 
acts  downward.  Finally,  P  =  V{io.6f+  1.14^)  =  10.73  tons,  and  the  in- 
clination of  P  with  the  horizontal  is  tan^^  (i-i4  "^  10.67)  =  6°  7';  and  now 
all  the  forces  on  each  member  are  determined,  those  on  the  post  being 
represented  in  the  figure. 

Generally,  several  sketches  may  be  made  and  considered  in  several  differ- 
ent orders,  each  furnishing  a  complete  analysis.  For  example,  we  might  have 
taken  the  entire  crane,  the  boom,  and  the  post;  or  the  brace,  the  boom,  and 
the  entire  crane.  The  student  is  advised  to  try  these  orders  and  make  the 
analysis. 

The  graphic  method  of  solving  the  various  force  systems  may  be  carried  out 
as  follows:  The  system  acting  on  the  entire  crane  consists  of  four  forces,  and 
so  the  resultant  of  any  pair  of  the  four  forces,  as  Nx  and  Ny,  balances  the  other 
pair;  therefore  that  resultant  is  concurrent  with  the  second  pair  and  acts  in 
the  line  1-2  (Fig.  105).     So  we  draw  the  force  triangle  ABC  A  for  those  three 


forces  (making  AB  represent  8  tons),  and  find  that  BC  represents  M  and  CA 
the  resultant  of  the  first  pair.     Next  we  resolve  CA  into  components  parallel 

to  Nx  and  Ny,  and  find  that  CD  and 


.  Ceiling 


DA  represent  Ny  and  Nx  respectively. 
The  forces  on  the  boom  being  three  in 
number  (the  load,  Q,  and  P),  they  must 
be  parallel  or  concurrent,  and  because 
two  (the  load  and  Q)  are  concurrent, 
all  must  be;  thus  the  line  of  action  of 
P  is  determined.  So  we  may  draw  the 
force  triangle  EFGE  for  the  three  forces, 
W^^p^///////yy////////^^^^^  making  EF  represent  8  tons;  thus  we 

find  that  EG  =  P  and  GF  =  Q. 

Example  (ii). — For  another  illus- 
tration, we  analyze  the  hydraulic  crane  represented  in  Figure  106.  It  consists 
of  a  hollow  post  MN  (up  into  which  the  piston  can  be  projected)  a  boom  PQ, 


Floor 


Fig.  106 


\rt.  17 


67 


and  a  pin-connected  frame  KPQ.  A  single  roller  is  mounted  on  the  pin  K, 
and  two  on  the  pin  P,  so  that  as  the  piston  moves  the  frame  moves  with  it, 
all  rollers  rolling  on  the  post.  Thus  there  are  twelve  parts:  a  post,  a  boom, 
two  struts  KP  (one  on  each  side  of  the  post),  two  ties  KQ  (one  on  each  side), 
a  pin  at  P,  one  at  Q,  one  at  K,  two  rollers  at  P  and  one  at  K.  We  take  the 
load  as  10  tons  and  x=  15  feet,  and  neglect  the  weights  of  the  parts. 

Fig.  107  represents  the  entire  crane,  not 
including  the  piston,  with  aU  the  external 
forces  acting  upon  it.  2Fx  =  o  shows  that 
M  =  Nx,  and  SMat  =  o  shows  that  M  = 
(10X15)-^^  where  h  =  height  of  post.  L  and 
iV„  cannot  be  found  from  this  force  system;  so 
we  try  the  frame  with  rollers  (Fig.  108) .  The 
external  forces  acting  on  it  are  the  load,  the 
piston  pressure  L,  the  post  pressure  Ri  against  the  single  roller,  and  the  result- 


^'M 

\ 

1^"^- 

.,  Q 

V) 

— •! 

1 

1. 

10^ 

tons 

h\ 


Fig.  107 


Py 


=3 


lojtons  lOitons 

Fig.  109 


ant  post  pressure  Ro  against  the  lower  rollers.  The  solution  of  this  system 
falls  under  Art.  13;  it  shows  that  L  =  10  and  Ri  =  R2=  21.4  tons.  Fig.  109 
represents  the  boom  alone  and  the  external  forces  acting  upon  it,  —  the  load, 
the  piston  pressure,  the  pin  pressure  Q  (acting  along  the  ties  because  each  is 
a  two-force  member),  and  the  pin  pressure  P  whose  direction  is  unknown. 
The  solution  of  this  system  falls  under  Art.  13.  DeaUng  with  the  unknown 
components  of  P  (senses  guessed  at),  we  find  from  XMp  =  o  that  Q  =  25.2 
tons;  from  SF^  =  o  that  Px  =  24  tons;  and  from  SF^  =  o  that  Py=  7.2 
tons.     The  pin  at  K  (Fig.  no)  is  subjected  to  three  forces,  namely,  the  pull  of 

the  two  ties  (25.2  tons),  the  pressure 
of  the  roller  (21.4),  and  the  force  F 
exerted  by  the  struts  (along  the  axis 
of  the  struts,  since  each  is  a  two- 
force  member).  From  2F„  =  o  we 
find  that  F  =  7.8  tons.  Fig.  no 
also  represents  the  post  with  all 
Since  2F„  =  o,  iV„  =  o;  SMtv  =  o  gives 
gives  Nx  =  M.    Now  we  have  all  the 


fr<l.        21.4 


w  tons 


K 

Z\At 

N, 


■M 


K 


21.4 1. 


# 

21.4  \      ^ 


Tn 


Fig. 


y-V 
IIO 


the  external  forces  actmg  upon  it. 

M  =  149.8  ^  MN;    and  SF^  =  o 

forces  on  each  part;  each  tie  is  subjected  to  end  pulls  of  12.6  tons;  each  strut 

to  end  pushes  of  3.9  tons;  the  pin  at  P  to  three  forces  as  shown  in  the  figure. 


68 


Chap,  hi 


The  graphical  solutions  of  the  various  force  systems  might  be  carried  out 
as  follows:  Four  forces  act  on  the  portion  of  the  crane  shown  in  Fig.  in, — the 
load  lo  tons,  the  pressures  L,  Ri,  and  R^.  The  resultant  R  oi  L  and  R2  acts 
through  their  intersection  and  through  that  of  Ri  and  the  load,  hence  in  the 
line  1-2.  The  load,  Ri,  and  R  are  in  equilibrium;  so  we  draw  a  closed  force 
polygon  for  them  as  ABC  A  (Fig.  112);  AB  =  10  tons,  BC  =  21.4,  and  CA 


lOYtons 


Fig.  Ill 


Fig.  112 


represents  R.  Finally  we  resolve  R  into  its  two  components;  CD  and  DA 
represent  L  and  R2  respectively.  There  are  four  forces  acting  on  the  boom, 
namely,  the  load  =  10  tons,  L  =  10  tons,  the  pin  pressure  P,  and  that  at  Q 
(Fig.  113).  Obviously  the  pressure  Q  acts  along  the  tie  rod.  The  first  pair  of 
forces  named  constitute  a  couple;  and  since  a  couple  can  be  balanced  only 
by  another  couple,  the  second  pair  is  a  couple  and  P  is  parallel  to  Q,  and  the 
resultant  of  each  pair  therefore  acts  in  the  line  1-2.  We  now  draw  a  line 
through  B  (Fig.  112)  parallel  to  Q,  and  one  through  A  parallel  to  1-2;  then  BE 
represents  Q  and  AE  represents  the  resultant  of  L  and  P.  Finally,  there  are 
three  forces  acting  on  the  pin  at  P,  namely,  R2  (or  CB),  —P  (or  BE),  and 
the  pressure  of  the  braces  KP  (Fig.  in).  These  three  forces  being  on  equi- 
librium, the  last  one  is  represented  by  EC. 

Example  (iii).  — We  now  make  an  analysis  of  a  crane  taking  into  account 
the  weights  of  the  members.  For  this  purpose  we  take  the  crane  described 
in  example  (i)  and  assume  that  the  weights  of  members  are  as  follows:  MN  = 
0.8  ton,  PQ  =  0.9  ton,  and  KQ  =1.1  tons.  The  load  is  taken,  as  in  example  (i), 
to  be  8  tons  at  16  feet  out  from  the  axis  of  the  post,  and  the  boom  22  feet  long. 
Fig.  114  shows  the  entire  crane  and  all  the  external  forces  acting  upon  it 
so  far  as  known.     Determination  of  the  unknown  reactions  M,  N^,  and  Ny 

presents  typical  problem  (v)  (Art. 
8.09ton5  J3)_  YTom  ^AIm  =  o  we  get 
iVx  =  8.09;  from  2F^  =  0,  M  = 
8.09;  and  from  2Fy  =  0,  Ny  = 
10.8.  Fig.  115  represents  the 
post  and  all  the  external  forces 
acting  upon  it  so  far  as  known. 
The  pressures  on  the  post  are 
exerted  by  members  which  are 
not  two  force  members,  and 
therefore  those  pressures  do  not 


0.9  tons 


tons 


M 


8.09^ 
tons 


'0.8  tons 


K 


N 
lO.O&tons 


Fig.  114 


Fig.  115 

act  in  the  directions  of  the  boom  and  brace.     The  directions  of  those  pres- 
sures being  unknown,  we  represent  each  by  its  (unknown)  horizontal  and 


Art.  1 8 


69 


vertical  component.  The  force  system  acting  on  the  post  contains  four 
unknowns,  namely,  Px,  Py,  K^,  and  Ky.  Not  all  of  these  unknowns  can  be 
determined  from  a  study  of  this  system  alone;  but  two  of  them,  Px  and  Kx, 
can  be  so  determined.  'LMk  =  o  gives  Px  =  12.13,  ^^^  ^^x  —  o  gives  Kx  = 
12.13  tons. 

Fig.  116  shows  the  boom  and  the  forces  acting  upon  it  so  far  as  known. 
The  direction  of  the  pressure  at  Q  is  unknown  as  yet;  therefore  that  pressure 
is  represented  by  means  of  its  (unknown)  components.     Determination  of  the 


pi 


12.13 
tons 


0.9  tons 


Gl 


6 1  tons 


10.95 
tons 


Fig.  116 


l2.l3ton5< 


->l2.13ton& 
Fig.  117 


unknowns  in  the  force  system  presents  typical  problem  (v).  2Fx  =  o  gives 
Qx  =  12.13  tons;  SA/q  =  o  gives  Py  =  0.95;  and  2F„  =  o  gives  Qy  =  9.85. 
Ha\ing  found  the  value  of  Py,  we  find  from  ZFy  =  o  for  Fig.  115  that  Ky  = 
10.95  tons.  To  check  the  analysis,  we  might  supply  values  of  the  forces 
acting  on  the  brace  (Fig.  117),  and  then  test  whether  the  force  system  is 
balanced,  that  is,  whether  S/^i  =  o,  XFy  =  o,  and  2M  =  o. 


18.   Cranes.  —  Continued 

In  this  article  we  show  how  to  analyze  three  cranes,  paying  some  attention 
to  the  forces  due  to  the  hoisting  rig.  Generally,  a  pulley  is  an  important  part 
of  such  rig.  We  assume  here  that  the  tensions  Ti  and  T2  (Fig.  118)  in  the  rope 
or  chain  on  opposite  sides  of  the  pulley  on  which  it  bears 
are  equal.  This  assumption  impHes  perfect  flexibility  of 
rope  or  chain  and  a  frictionless  pin  supporting  the  pulley. 
The  pressure  P  against  the  pin  equals  the  resultant  of 
those  tensions,  or  2  T  cos  ^  a,  and  it  bisects  the  angle 
between  their  lines  of  action.  If  the  Hnes  of  action  are 
parallel  (a  =  o),  P  =  2  T;  if  they  are  at  right  angles 
{a  =  90°),  P  =  1.414  T, 

Example  (i).  — Fig.  119  represents  a  crane  supported  in  a  footstep  bearing 
at  the  floor  and  a  collar  bearing  on  the  wall  bracket  H.  The  hoisting  rig  con- 
sists of  a  simple  hand  winch  mounted  on  the  wall  at  W,  a  chain,  and  pulleys 
as  shown.  Pulley  at  G  is  12  inches  in  diameter;  the  load  is  one-half  ton. 
The  reactions  at  the  supports  depend  on  the  hoisting  rig,  as  will  be  seen  from 
the  following:  On  the  entire  crane,  including  the  top  pulley  (Fig.  120),  there 
are  acting  four  forces,  namely,  the  upper  reaction  U,  the  lower  reactions  P, 


Fig 


70 


Chap,  iir 


and  Py,  and  the  pressure  of  the  chain  against  the  pulley  equivalent  to  two 
components,  one-half  ton  each,  as  shown.  Taking  moments  about  the  lower 
end,  we  find  H  to  be  0.0S7  ton;  from  SF^  =  o  and  SFj,  =  o,  we  find  that 


5i       K6=I2' 
KJ  =  5' 
PKJ  =6KJ 


Fig.  119 


Fig.  120 


Px  =  0.413  and  Py  =  0.5  ton.  All  members  except  the  vertical  HP  are  simple 
tension  or  compression  members.  Force  polygons  for  joints  G  and  /  show 
that  the  stresses  are  as  follows:  GK  =  0.35  ton  (tension);  GJ  =  1  ton  (com- 
pression); JK  =  0.57  ton  (compression);  JP  =  i  ton  (compression).  Mem- 
ber HP  is  subjected  to  the  reactions  of  the  supports  as  already  computed,  and 
the  following  forces:  a  pull  of  0.35  ton  along  KG;  a  push  of  0.57  ton  along  KJ; 
and  a  push  of  i  ton  along  PJ. 

Example  (ii).  —  Fig.  121  represents  a  common  type  of  derrick.     It  is  sup- 
ported by  a  footstep  at  the  bottom  of  post  and  at  the  top  by  two  stiff  legs 


""'^W////////////////////'- 


Fig. 


121 


■A 


>- 


^2T 


which  extend  backward  to  the  ground  or  other  base;  the  spread  (angle  be- 
tween their  horizontal  projections)  being  90  degrees  so  that  the  derrick  can 
swing  about  its  vertical  axis  through  270  degrees.     Sometimes  the  derrick  is 


Art.  i8  yi 

supported  at  the  top  by  a  collar  bearing  held  in  place  by  cables  extending  off 
to  quite  remote  points  on  the  ground. 

Obviously  the  pull  on  a  stiff  leg  is  greatest  when  the  boom  is  in  the  same 
plane  with  that  leg;  the  pull  on  a  cable  is  greatest  when  that  cable  and  the 
boom  are  in  the  same  plane  and  on  opposite  sides  of  the  post.  Let  P  denote 
this  pull,  and  a  the  inclination  of  the  cable  to  the  horizontal  or  the  inclination 
of  the  line  joining  the  pivot  on  the  post  with  the  lower  end  of  a  stiff  leg.  Then 
taking  moments  of  all  external  forces  on  the  derrick  about  the  footstep  bear- 
ing, we  get  Ph  cos  a  =  Ws,  or  P  =  Ws/h  cos  a  (only  the  weight  of  the  load 
being  taken  into  account).  Calling  the  horizontal  and  the  vertical  reactions 
at  the  footstep  H  and  V  respectively,  we  find  that  H  =  Ws/h  and  V  =  W  -{- 
P  sin  a  =  W(i  +  tan  a  •  s/h). 

There  are  seven  forces  acting  on  the  part  shown  in  Fig.  122,  which  consists 
of  the  crane  post,  the  winch  W,  the  two  sheaves  S,  and  a  part  of  the  hoisting 
and  topping  ropes  as  shown.  The  forces  are:  H,  V,  and  P  (already  explained) ; 
Q,  the  pressure  of  the  boom  on  the  post  acting  in  a  direction  as  yet  unknown; 
^  W,  approximate  value  of  the  tension  in  the  hoisting  rope;  T,  which  denotes 
the  tension  in  the  topping  rope;  and  2  T,  exerted  by  the  top  pulley  shackle. 
Of  these  seven  forces,  all  except  Q  and  T  are  already  known.  To  find  these 
we  may  proceed  as  follows :  Take  moments  of  all  the  forces  about  the  pin  at  Q, 
and  thus  find  T;  then  take  horizontal  and  vertical  components,  and  thus 
find  the  horizontal  and  vertical  components  of  Q,  and  finally  Q  itself.  The 
force  system  can  be  solved  graphically  as  follows :  First  find  the  line  of  action 
of  the  resultant  R  of  the  two  forces  T  and  2  T;  then  this  R  and  the  other  five 
forces  constitute  a  system  in  equilibrium,  which  solve  for  R  and  Q  by  methods 
explained  in  Art.  13;  finally  resolve  R  into  its  components  T  and  2  T. 

Example  (iii).  —  Fig.  123  represents  a  sheer  leg  crane.  It  consists  of  two 
front  legs  AC  and  BC  and  a  back  leg  CD,  all  connected  by  a  horizontal  pin 
at  C;   the  front  legs  are  pin-supported  on 

the  ground  at  A  and  B,  and  the  back  leg  ,,^^jf 

is  restrained  at  the  ground  by  a  holding-  .^^^/^  h 

down  rail  and  a  long  horizontal  screw  which  ^.f^^y  //        I 

works  in  a  nut  on  the  lower  end  D.    The  ^^^  ^'^'^-^^-.^.^     I ■■ 

purpose  of  the  screw  is  to  move  D,   thus       .^^-—'^ ^^^^=s=^.. 

turning  the  front  legs  about  AB  and  moving  '•=^=-~  ^--^ 

the  load  in  and  out.     We  will  now  show  how 

to  determine  the  pressures  on  the  ends  of  the  legs  due  to  their  own  weights, 
taking  the  following  data:  lengths  of  front  legs  160  feet,  distance  between 
their  lower  ends  50  feet,  distance  between  their  upper  ends  10  feet,  length 
of  back  stay  210  feet,  weight  of  each  front  leg  44  tons,  of  the  back  leg  53 
tons;  we  take  the  crane  in  its  position  of  greatest  overhang,  64  feet. 

The  external  forces  acting  on  the  crane  are  the  following  (see  Fig.  124): 
the  three  weights,  the  holding-down  force  Dy,  the  push  of  the  screw  Dx,  the 
inward  pushes  At  and  Bz  of  the  supports  at  A  and  B,  and  the  pressures  of  the 


_.  Chap,  ui 

72 

pins  at  A  and  B;  each  of  these  pressures  is  represented  by  two  components 
in  the  figure,  A^,  Ay,  and  B^,  By,  respectively.  There  are  six  conditions  of 
equilibrium  for  this  system,  namely,  the  sums  of  the  components  of  the  forces 
along  the  x,  y,  and  z  axes,  and  the  sums  of  the  moments  about  those  axes 
equal  zero.     Thus,  — 


ZFy=    Ay+By-    Dy-     SS-44-    44=0 

XF,  =  -  A,-\-B^  =  o 

•ZM^=  -Ay  X  25  +  J5^  X  25  +  44  X  15  -  44  X  15  =  o 

2M„=^xX  25-5xX  25  =  0 

S7lf.=  I>„X87.6  +  53X  11.8-44X32X  2  =  0 


(i) 
(2) 
(3) 
(4) 
(5) 
(6) 


Equation  (6)  shows  that  Dj,  =  25  tons;  (4)  shows  that  Ay  =  By-,  from  these 
results  and  (2)  it  follows  that  Ay  and  By  equal  83  tons.  No  other  unknowns 
can  be  determined  from  the  equations;  but  (3)  shows  that  Az  =  B^,  (5)  that 
A,  =  Bx,  and  (i)  that  A^-\-B^^D^. 


V^ 


44- tons 


ze^-^^l- 


1 


75.8:.^ 75.8'--^   ^^,  l>^r;%7"^' 

tons 


Fig.  125 


Fig.  126 


To  get  values  of  these  unknowns  we  consider  the  forces  acting  on  the  back 
leg;  there  are  four  forces,  namely,  the  weight  of  the  leg  (53  tons),  the  holding- 
down  force  Dy  (25  tons),  the  screw  pressure  Dz,  and  the  pressure  of  the  upper 
pin  at  C,  represented  for  convenience  by  two  components  which  we  call  Cx  and 
Cy  (Fig.  125).  This  system  is  in  equilibrium  and  so  Slfc  =  25  X  151. 6  — 
D^  X  145-2  +  53  X  75-8  =  o,  or  D^  =  53.8  tons;  SF^  =  C^  -  53.8  =  o,  or 
^'  —  53-8;  and  I^Fy  =  Cj,  —  25  —  53  =  o,  or  Cy  =  78.  Returning  now  to 
equations  (i)  and  (5),  we  find  that  Ax  and  Bx  =  26.9.  To  get  Az  and  B^  it  is 
necessary  to  discuss  the  forces  on  one  of  the  front  legs.  There  are  three 
forces,  —  the  weight  44  tons,  and  the  pressures  at  the  ends;  each  of  the  pres- 
sures is  represented  (Fig.  126)  by  three  components,  26.9,  83,  and  B^  below, 
and  Qx,  Qy  and  Qz  above.  The  system  being  in  equilibrium,  we  take  moments 
about  the  vertical  line  through  Q;  thus  5^  X  64  —  26.9  X  20  =  o,  or  B^  = 
8.41  tons.    Inspection  shows  that  Qx  =  26.9,  Qy  =  39,  and  Qz  =  8.41  tons. 


Art.  1 8 


73 


lb. 2  tons 


|<- 75.8'-  ■  ->) ;  K-J2'>H-32'-i 


The  forces  acting  on  the  upper  pin  (at  C)  are  represented  in  Fig.  127,  by 
means  of  their  components. 

We  now  give  another  solution  of  the  foregoing  example, 
making  use  of  the  principle  that  if  the  forces  of  a  system 

i  n  equilibrium 
b  e  represented 
by  vectors,  then 
the  projection  of 
those  vectors  on  Fig.  127 

any  plane  represents  a  force  system 
also  in  equilibrium  (Art.  10  under 
(vii)).  Projecting  the  force  system 
represented  in  Fig.  124  on  the  three 
coordinate  planes,  we  get  the  three 
systems  represented  in  Fig.  128, — 
side  elevation,  end  elevation,  and 
plan.  From  the  side  elevation, 
^Ma  =  o  gives  Dy=  2%  tons; 
2i//>  =  o  shows  that  Ay  =  By] 
and  ^Fy  shows  that  Ay-\-  By  = 
166,  or  ^1,  and  By  =  83  tons.  No  further  numerical  result  can  be  obtained 
from  these  projected  systems.  Considering  the  back  leg  alone  as  before,  we 
would  find  that  Z)^  =  53-8  tons;  then  from  the  plan  A^=  Bx  obviously, 
and  Ax+Bx=  53.8,  or  A^  and  -Sx  =  26.9  tons.  Az  and  Bz  would  be 
gotten  as  before.* 

*  For  full  information  on  cranes,  see  Bottcher's  book  on  that  subject,  English  translation 
by  Tolhausen. 


Side  Elevation 


53.8ton5 

* — _ 


[<-?5''>{<25'>l 

End 
Elevation 


CHAPTER    IV 


FRICTION 


19.   Definitions  and  General  Principles 

§  I.  Definitions,  Etc.  —  When  one  body  slides  or  tends  to  slide  over  an- 
other, then  the  sliding  of  the  first  or  its  tendency  to  slide  is  resisted  by  the 
second.     Thus,,  if  A  (Fig.  129)  is  a  body  which  slides  or  tends  to  slide  toward 

the  right  over  B,  then  B  is  exerting  some  such  force 
as  i?  on  ^,  and  the  component  of  R  along  the  surface 
of  contact  is  the  resistance  which  B  offers  to  the  sliding 
or  tendency.  Of  course  A  exerts  on  5  a  force  equal 
and  opposite  to  R;  either  of  these  equal  forces  is 
called  the  total  reaction  between  the  two  bodies.  The 
component  of  either  total  reaction  along  the  (plane)  surface  of  contact  is 
called  friction,  and  the  component  of  either  along  the  normal  is  called  normal 
pressure;  they  will  be  denoted  by  F  and  N  respectively.  If  the  surface  of 
contact  of  the  two  bodies  is  not  plane,  the  force  exerted  at  each  elementary 
part  of  the  surface  is  the  total  reaction  at  that  element,  and  its  components  in 
and  normal  to  the  element  are  the  friction  and  the  normal  pressure  at  the  ele- 
ment. Friction  is  called  kinetic  or  static  according  as  sliding  does  or  does  not 
take  place.     Only  static  friction  is  considered  here. 


'W/////^//M^///M/////'//////' 


Fig.  129 


W////////////.  '///.^////////A        •'''//////////////y,  V///.ii'///////A 


'5 


YW 


4  lbs. 


'■^6' 


YW 

Fig.  130 


YW 


The  amount  of  static  friction  between  two  bodies  depends  upon  the  degree 
of  the  tendency  to  slip.  Thus  suppose  that  A  (Fig.  130)  is  a  block  weighing 
10  pounds,  upon  a  horizontal  surface  B\  that  the  block  is  subjected  to  a  hori- 
zontal pull  P,  and  that  the  pull  must  exceed  6  pounds  to  start  the  block. 
Obviously  when  P  =  2  pounds  say,  then  F  =  2;  when  P  =  4  pounds,  then 
F  =  4;  etc.,  until  motion  begins.  So  long  as  P  does  not  exceed  6  pounds,  F 
equals  P;  that  is,  F  is  passive  and  changes  just  as  P  changes.  The  inclination 
of  the  reaction  R  also  depends  on  the  degree  of  the  tendency  to  slip.  When 
P  =  2  pounds,  then  the  angle  NOR  =  tan-^  f^  =  11°  19';  when  P  =  4  pounds, 

74 


Art.  19  75 

NOR  =  tan-'  t%  =21°  48';  etc.,  until  motion  begins.  The  greatest  values  of 
the  friction  F  and  the  angle  NOR  obtain  when  motion  impends. 

The  friction  corresponding  to  impending  motion  is  called  limiting  friction. 
We  wnll  denote  it  by  Fm,  since  it  is  a  maximum  value  (see  Fig.  130).  The 
coefficient  of  static  friction  for  two  surfaces  is  the  ratio  of  the  limiting  friction 
corresponding  to  any  normal  pressure  between  the  surfaces  and  that  normal 
pressure.     We  will  denote  it  by  /i;   then 

M  =  FJN,  or  F„,  =  ixN;  also,  F  >  nN. 

The  angle  of  friction  for  two  surfaces  is  the  angle  between  the  directions  of  the 
normal  pressure  and  the  total  reaction  when  motion  is  impending.  We  will 
denote  it  by  0  (see  Fig.  130);    then 

tan<^  =  Fm/N;  hence  tan</>  =  n. 

If  a  block  were  placed  upon  an  inclined  plane,  the  inclination  at  which  slipping 

would  impend  is  called  the  angle  of  repose  for  the  two  rubbing  surfaces;  it  will 

be  denoted  by  p.     The  angles  of  friction  and  repose  for 

two  surfaces  are  equal;  proof  follows:   Suppose  that  A  \^ 

(Fig.  131)  is  on  the  point  of  sliding  down  the  incline; 

two  forces  act  on  A ,  its  own  weight  W  and  the  reaction        ^^ 

R  of  the  plane.    Since  A  is  at  rest,  R  and  W  are  colinear,        \ffff^ 

that  is,  R  is  vertical;  and  since  motion  impends,  the     <0--^ 

angle  between  R  and  the  normal  is  the  angle  of  fric-  ^^^-  ^^^ 

tion  (/).     It  follows,  from  the  geometry  of  the  figure,  that  0  and  p  are  equal. 

The  coefficient  of  static  friction  for  two  bodies  A  and  B  may  be  found  in 
several  ways:  (i)  Place  ^  on  5  as  in  Fig.  130,  and  determine  the  pull  P  which 
will  just  start  A;  then  fx  =  P  divided  by  the  weight  of  ^.  Or  (ii)  tilt  B,  and 
determine  the  inclination  at  which  gravity  will  start  A  down;  then  ju  equals 
the  tangent  of  that  angle  of  inclination.  In  either  method  several  determina- 
tions must  be  made  to  obtain  a  fair  average.  Many  experiments  have  been 
made  in  these  ways,  and  it  has  been  ascertained  that  coefficients  of  static 
friction  depend  on  the  nature  of  the  materials,  character  of  rubbing  surfaces 
and  kind  of  lubricant,  if  any  be  used.  Early  experimenters  reported  (Coulomb 
1871,  Rennie  1828,  Morin  1834,  and  others)  that  the  coefficient  is  independent 
of  the  intensity  of  normal  pressure;  and  although  this  announcement  was 
clearly  subject  to  the  limitation  of  the  range  of  the  experiments  performed, 
yet  it  was  generalized  and  long  accepted  as  a  universal  law  of  friction.  But 
the  universality  of  the  law  has  been  questioned;  Morin  himself  pointed  out 
that  length  of  time  of  contact  of  the  two  bodies  influences  the  coefficient;  and 
obviously  the  coefficient  changes  when  the  intensities  of  pressure  get  so  low 
that  a  considerable  part  of  the  friction  is  due  to  adhesion,  or  so  high  as  to  affect 
the  character  of  the  surfaces  in  contact.     Messiter  and  Hanson  report*  prac- 

*  Eng.  News,  1895,  Vol.  33,  page  322. 


^  Chap,  iv 

76 

tical  constancy  of  coefficient  for  yellow  pine  and  spruce.    They  give  the 
following  for  planed  or  sandpapered  (i)  yellow  pine  and  (2)  spruce. 

(i)  n  =  0.25  to  0.32;  average  M  =  0.29  for  100  to  1000  lbs.  per  sq.  in. 

(2)  M  =  0.18  to  0.53;  average  M  =  0.42  for  100  to  i6oolbs.per  sq.  in. 

The  variation  depends  on  relation  of  grain  of  wood  to  direction  of  slide. 

Coefficients  of  Static  Friction 

(Compiled  by  Rankine  from  experiments  by  Morin  and  others.) 

Dry  masonry  and  brickwork o-6    to  0.7 

Masonry  and  brickwork  with  damp  mortar 0.74 

Timber  on  stone about    0.4 

Iron  on  stone ^-3    to  0.7 

Timber  on  timber o-2    to  0.5 

Timber  on  metals o-2    to  0.6 

Metals  on  metals o-i5  to  0.25 

Masonry  on  dry  clay o-S^ 

Masonry  on  moist  clay ^-33 

Earth  on  earth 0-25  to  i.o 

Earth  on  earth,  dry  sand,  clay,  and  mixed  earth 0.38  to  0.75 

Earth  on  earth,  damp  clay i-o 

Earth  on  earth,  wet  clay 0.31 

Earth  on  earth,  shingle  and  gravel 0.81  to  i.ii 

§  2.  Tractive  Force.  —  Let  W  =  the  weight  of  a  body  A  upon  a  horizontal 
surface  B  (Fig.  132),  m  =  the  coefficient  of  friction  for  the  surfaces  in  contact, 
<f)  =  their  angle  of  friction,  and  P  a  force  applied  to  the  body  as  shown,  6  being 
the  inclination  of  P  to  the  horizontal.  Then  the  force  P  required  to  start  the 
body  to  move  is  given  by 

fxW  Wsm(f> 


P  = 


cos  d  -\-  lisind      cos  {d  —  <f>) 


-.6 


Fig.  132 


4 


A   V 


e 


w 


Fig.  133 


Fig.  134 


Fig.  13s 


The  forces  acting  on  A  are  P,  W,  and  the  reaction  of  the  plane  whose  two 
components  are  A''  and  (when  motion  impends)  Fm  (Fig.  133).  Now  P  cos  Q  = 
F„,  iV  =  PF  —  Psin0,  and  Fm  =  m^V;  these  three  equations  solved  simul- 
taneously furnish  the  first  stated  value  of  P.  The  second  value  can  be  ob- 
tained from  the  first,  or  by  solving  the  three-force  system  acting  on  A  as  repre- 
sented in  Fig.  134.  According  to  Lami's  theorem  (Art.  10),  P/PF  =  sin  0/sin 
(90  -f  0  -  0);  hence  P  =  M^  sin  0/cos  (6  -  (p). 


Art.  19 


77 


If  the  pull  P  is  horizontal  (d  =  o),  then  P  =  /iW.  If  the  pull  is  inclined, 
but  not  too  much,  then  the  pull  P  required  to  start  the  body  may  be  less  than 
nW.  In  fact  the  least  value  of  P  obtains  when  6  =  (f),  —  "the  best  angle  of 
traction  equals  the  angle  of  friction,"  —  and  the  minimum  value  of  the  pull 
is  Wsincj).  Proofs  follow:  (i)  Evidently  W  sincf)  -r-  cos  (0  —  0),  the  general 
value  of  P,  changes  as  9  changes,  and,  for  a  given  W  and  <f),  P  is  least  when 
cos  (6  —  (j))  is  greatest;  but  this  greatest  value  is  i,  and  obtains  when  5  —  0  =  o, 
or  when  0  =  </>  as  stated,  etc.  (ii)  Or,  let  AB  (Fig.  135)  represent  W,  BC  be 
parallel  to  P,  and  AC  he  parallel  to  R;  then  CA  represents  R.  If  0  be  changed, 
then  BC  (and  P)  will  change;  and  evidently  P  will  be  least  when  BC  is  per- 
pendicular to  CA,  that  is,  when  d  =  (j).     And  then  BC  (or  P)  =  W  sin  0. 

§  3.  Test  for  Rest  or  Motion.  —  A  body  is  supported  so  that  it  can  slip 
and  is  subjected  to  given  forces;  it  is  required  to  ascertain  whether  those  forces 
do  cause  slipping,  and  the  value  of  the  friction  is  desired.  We  assume  that 
the  body  is  at  rest,  and  determine  the  friction  F  and  the  normal  pressure  N 
from  conditions  or  equations  of  equilibrium;  then  we  compare  F  with  fxN. 
If  F  is  less  than  /jlN,  there  is  no  motion  and  the  computed  value  of  F  is  correct; 
if  F  is  greater  than  fiN,  then  there  is  motion  and  the  friction  is  kinetic,  its 
value  being  less  than  jjlN.  For  example,  consider  a  block  of  material  weighing 
100  pounds  supported  on  a  horizontal  surface,  the  coeflEicient  of  friction  being 
^,  and  imagine  a  down  push  of  200  pounds  applied  to  the  block  at  an  angle  of 
30  degrees  with  the  vertical.  N  =  100  +  200  cos  30  =  273.2,  and  for  rest, 
F  =  200  sin  30  =  100;  ijlN  =  I  X  273.2  =  136.6,  and  this  is  the  greatest  fric- 
tional  resistance  which  the  support  can  offer  so  long  as  N  =  273.2.  Only 
100  pounds  are  required  to  prevent  motion,  and  so  the  body  is  at  rest  under 
the  action  of  friction  of  that  required  value. 

Or,  to  test  for  rest  or  motion,  we  may  make  use  of  the  so-called  cone  of  jric- 
tion  for  the  two  bodies  in  contact,  which  may  be  described  thus:  Let  P  (Fig. 
136  or  137)  denote  the  resultant  of  all  the  forces 
applied  to  or  acting  on  the  body  A  (whose 
state  is  to  be  investigated)  but  not  including 
the  total  reaction  of  the  supporting  body  B; 
0  the  point  where  P  cuts  the  surface  of  con- 
tact between  A  and  B,  and  DOC  equal  the 
angle  of  friction;  then  the  cone  generated  by 
revolving  OC  about  OD  is  the  cone  of  friction. 
If  the  line  of  action  of  the  resultant  P  does 
not  fall  outside  the  cone  (Fig.  136),  then  there 
is  no  slipping;  if  it  does  fall  outside  (Fig.  137), 
then  there  is  slipping.  Proof  follows:  As 
already  pointed  out,  the  direction  of  the  total  reaction  i?  on  a  body,  which 
tends  to  slide  over  another,  depends  on  the  degree  of  the  tendency;  the  greater 
the  tendency,  the  greater  the  inclination  of  R  from  the  normal;  but  the  in- 
clination has  a  limit,  that  limit  being  equal  to  the  angle  of  friction,  and  it 


^o  Chap,  iv 

obtains  when  slipping  impends.  Therefore  when  P  acts  within  the  cone  or 
along  an  element  of  it,  then  R  can  incline  and  completely  oppose  P  (Fig.  136), 
no  matter  how  large  P  may  be.  When  P  falls  outside  the  cone,  R  can  incline 
only  to  an  element,  and  the  friction  cannot  successfully  oppose  the  component 
of  P  which  tends  to  move  the  body  (Fig.  137).  In  the  preceding  example  P 
is  the  resultant  of  the  weight  of  the  block  100  pounds,  and  the  applied  push 
200  pounds.  That  resultant  makes  an  angle  of  10°  t,2>'  '^^'ith  W  or  the  normal. 
The  angle  of  friction  is  tan"^  i  or  26°  34';  hence  P  falls  inside  the  cone  and, 
according  to  the  principle  of  the  cone,  motion  does  not  ensue. 

As  another  application  of  the  cone  principle  consider  Fig.  138,  which  repre- 
sents (in  plan  and  elevation)  a  type  of  simple  hanger.     It  consists  of  a  fixed 

vertical  rod  and  a  horizontal  piece  which  is 
forked;  there  is  a  hole  in  each  part  of  the 
fork  so  that  the  piece  can  be  slipped  over  the 
rod  as  shown  in  the  elevation.  The  hanger, 
if  properly  made,  will  not  slip  down  along  the 
rod  on  account  of  its  own  weight  or  that  of  a 
load  unless  it  be  hung  quite  close  to  the  fork. 
The  mechanics  of  the  device  may  be  explained 
as  follows:  Obviously  the  rod  reacts  on  the 
hanger  at  Oi  and  O2.  When  slipping  impends 
at  these  points,  the  reactions  act  along  OiCi 
and  O2C2  inclined  to  the  normals  an  amount 
equal  to  the  angle  of  friction  0  as  shown. 
The  hanger  being  at  rest  (by  supposition),  the  third  force  acting  upon  it 
(the  load,  weight  of  hanger  neglected)  must  be  concurrent  with  these  two 
reactions;  hence  to  just  put  the  hanger  on  the  point  of  slipping,  the  load 
must  be  hung  from  a  point  in  the  vertical  through  C.  If  the  load  is  hung 
out  beyond  C,  as  at  A,  the  hanger  will  not  slip.  For  suppose  slipping  to 
impend  at  Oi,  then  R  at  Ox  would  act  along  OiCi,  and  R  and  W  would  concur 
at  a.  To  preserve  equilibrium,  R  at  O2  must  also  act  through  a,  which 
is  possible,  since  O^a  is  within  the  cone.  Or  suppose  slipping  to  impend  at 
O2,  then  R  at  O2  would  act  along  O2C2,  and  R  and  W  would  concur  at  m.  To 
preserve  equilibrium,  R  at  Oi  must  also  act  through  m  which  is  possible.  In 
similar  manner,  it  can  be  shown  that  a  load  hung  between  the  rod  and  C,  as  at 
B,  would  cause  slipping. 


"SP 


Fig.  138 


20.   Friction  in  Some  Mechanical  Devices 

§  I.  Inclined  Plane.  —  Let  a  =  the  inclination  of  the  plane  to  the  horizontal 
(Fig.  139),  p  =  angle  of  repose  for  the  plane  and  a  particular  body  upon  it, 
</)  =  their  angle  of  friction,  ^  =  coefficient  of  friction,  W  =  weight  of  the 
body,  and  6  =  angle  between  the  push  or  pull  P  and  the  incline,  (i)  The  pull 
P  required  to  start  the  body  up  the  plane  is  given  by 

Pi  =  W  sin  (a  +  0)/cos  {6-  -  4>) 


Art.  20 


79 


as  can  be  shown  by  means  of  Lami's  theorem  (Art.  10)  applied  to  the  three 
forces  acting  on  the  body  {P,  W,  and  the  reaction  R  of  the  plane).  Thus 
Pi/W  =  sin  (a  +  0)/sin  (90  —  </>  +  0) ;  hence,  etc.  Pi  is  a  minimum  (for  given 
W,  a,  and  </>)  when  6  =  <f>;  then  its  value  is  W  sin  (a  +  </>).  For,  it  is  ob- 
vious that  Pi  is  least  when  sin  (90  —  ^  +  6)  is  greatest,  that  is,  when  4>  =  d. 
(ii)  When  the  inclination  of  the  plane  is  greater  than  the  angle  of  repose 
(a  >  p  =  </)),  then  the  body  would  slip  down  unless  pre- 
vented by  a  suitable  force.  The  pull  P  required  to  prevent  the 
slipping  down  is  given  by 

P2  =  W  sin  {a  -  </))/cos  (0  +  <^). 

P2  is  a  minimum  when  6  =  —cf);  then  its   value  is  W  X 

sin  (a  —  4>).     (iii)  When  the  inclination  of  the  plane  is  less 

than  the  angle  of  repose  (a  <  p  =  <^),  then  the  body  would  not  slip  down 

on  account  of  its  own  weight.     The  push  P  required  to  start  the  body  down 

is  given  by 

P3  =  W  sin  (0  -  a)/cos  {(j>  +  d). 

Ps  is  a  minimum  when  6  =  —0;    then  its  value  is  W  sin  (</>  —  a). 

When  the  force  P  acts  along  the  plane  (d  =  o),  then  the  values  of  Pi,  P2,  and 
P3  are  respectively, 


Fig.  139 


W 


sin  (a  -\-  (/)) 

COS0 


W 


sin  (a  —  <f)) 

COS0 


W 


sin  ((f)  —  a) 

COS(f) 


§  2.  Wedge.  —  In  order  that  the  force  P  (Fig.  140)  may  start  the  wedge  in- 
ward to  overcome  the  load  W,  the  friction  at  the  three  rubbing  surfaces  must 
be  overcome  also.  If  the  three  rubbing  contacts  are  equally  rough  and  </>  = 
their  common  angle  of  friction,  then  the  force  P  required  to  start  the  wedge 
inward  is  given  by 

Pi  =  W  tan  (2  0  -f  a). 


wm/m/m^wm///////?///// 

Fig.  140 


Fig.  141 


Fig.  142 


Fig.  141  represents  the  three  forces  W,  Ri,  and  R2  acting  on  the  block  M;  also 
the  three  forces  R2"  (=  R2),  R3,  and  P  acting  on  the  wedge.  The  angles  which 
Ri,  R2,  and  R3  make  with  their  normal  components  equal  (p,  since  motion  im- 
pends, by  supposition.     In  Fig.  142,  ABCA  is  a  triangle  for  the  forces  acting 


8o 


Chap,  rv 


on  M,  AB  representing  W;  and  CBDC  is  a  triangle  for  the  forces  acting 
on  the  wedge.  The  given  formula  for  Pi  may  be  derived  from  these  triangles 
by  solving  for  BD,  which  represents  Pi.  From  the  first  triangle  (R2  =  R2") 
/W  =  cos  0/sin  (90  -  0  -  a  -  0),  or  R2  =  R2"  =  W  cos  0/cos  (2  0  +  a) ; 
from  the  second  triangle  Pi/{R2  =  R%")  =  sin  (2  0  +  a)/cos  0.  Therefore 
Pj  =  [R^'  =  R^")  sin  (2  0  +  a)/cos  0  =  IF  tan  (2  0  +  a). 

If  the  wedge  angle  a  is  less  than  2  0,  the  wedge  will  not  slip  out  under  any 
load  W  even  when  there  is  no  push  P;  that  is,  the  wedge  is  self-locking.  The 
force  required  to  pull  the  wedge  out,  that  is  to  lower  the  load  W,  must  equal 

W  tan  (2  0  —  a),  when      a  >  0  (guide  at  right  of  M), 


or 


W  sm  (2  0  —  a)  -T-  cos  a,     when      a  <  0  (guide  at  left  of  M). 

In  order  that  the  force  Q  (Fig.  143)  may  overcome  the  resistances  W,  the 
frictional  resistances  at  the  four  contacts  must  be  overcome  also.     If  the  con- 


FiG.  143 


Fig. 


144 


Fig.  14s 


tacts  are  equally  rough  and  0  =  their  common  angle  of  friction,  then  the  force 

^1 


necessary  to  start  the  wedge  down  is  given  by 

2W 


Fig.  146 


cot  (0  +  a:)  —  tan  0 

Fig.  144  represents  the  forces  Q,  R\ ,  and  R2  acting  on  the 
wedge,  and  the  forces  acting  on  M  and  N .  Each  of  the 
reactions  R  makes  with  its  normal  component  an  angle 
equal  to  0  (motion  impending).  In  Fig.  145,  ABCA  is 
a  triangle  for  the  forces  acting  on  i/,  AB  representing 
W .  AC  DA  is  a  triangle  for  the  forces  acting  on  the 
wedge.  The  given  formula  for  Qi  can  be  derived  from 
these  triangles  by  solving  them  for  DA,  which  represents 

Qu 

If  the  wedge  angle  2  a  is  less  than  2  0,  then  the  wedge 
would  not  slip  out  under  any  pressures  W  even  when  there 
is  no  push  Q;  that  is,  the  wedge  is  self-locking.  The  force 
required  to  pull  the  wedge  out  {M  and  N  guided  above)  is 
given  by 

0  '^^ 

^^      cot  (0  -  a)  +  tan  0 


Art.     20  ol 

§  3.  Screw.  —  Fig.  146  represents  a  simple  jackscrew  much  used  for  raising 
and  lowering  heavy  loads  through  short  distances.  In  the  simpler  forms, 
the  screw  is  turned  by  means  of  a  lever  stuck  through  a  hole  in  the  head  H 
of  the  screw.  There  is  frictional  resistance  between  the  screw  and  the  nut, 
also  between  the  cap  C  and  the  head  of  the  screw,  unless  the  load  can  turn 
with  the  screw.  Let  P  =  the  (horizontal)  force  applied  to  the  lever;  a  = 
the  arm  of  P  with  respect  to  the  axis  of  the  screw;  W  =  load  on  the  cap; 
r  =  mean  radius  of  the  screw,  |  (ri  +  ra);  a  =  pitch  angle  =  tan-^  (h  ^  2  7rr), 
where  h  =  pitch;  and  </>  =  tan-^,  where  m  =  coefiEicient  of  friction.  Dis- 
regarding the  friction  between  the  cap  and  head  of  the  screw,  the  moment 
required  to  raise  the  load  (or  move  the  screw  against  W)  is  given  by 

Pia  =  TFr  tan  (</)  +  «). 

If  the  pitch  angle  is  less  than  the  angle  of  friction,  the  load  would  not  turn  the 
screw;  that  is,  the  screw  is  self-locking.  The  moment  required  to  lower  the 
load  (or  move  the  screw  with  W)  is  given  by 

Pia  =  Wr  ta,n  (<^  —  a). 

Jackscrews  are  always  made  self -locking,  the  pitch  angle  a  being  between  4  and 
6  degrees  generally.     With  a  =  4  degrees  and  0  =  6  degrees  (m  =  o.i), 

PiC  =  0.18  Wr   and  P20  =  o-035  Wr. 

Derivation  of  formulas  for  Pi  and  P2:  At  each  point  of  contact  between 
the  screw  and  nut,  the  latter  exerts  a  pressure  dR  whose  normal  and  tangential 
component  we  call  dN  and  dF  respectively. 

(i)  When  the  screw  tends  to  rise,  dF  acts  downward  on  the  screw  as  shown 
at  A ;  and  when  motion  impends,  the  angle  between  dR  and  the  vertical  is 
(j)  -\-  a.  Taking  the  sum  of  the  vertical  components  of  all  the  forces  acting 
on  the  screw,  and  the  sum  of  the  moments  of  all  the  forces  about  the  axis 
of  the  screw,  we  get 

-W  +  ^  dR  cos  (<!>  -\-  a)  =  o,  or  cos  ((t>-\- a)'E  dR  =  W, 

and  Pia  —"EdR  sin  (</>  +  a)  r  =  o,  or  r  sin  ((^  -f  a)  2  (ii?  =  Pia. 

These  two  equations  combined  give  Pia  =  Wr  tan  (0  +  0;). 
(2)  When  the  screw  tends  to  descend,  dF  acts  upward  as  shown  at  B;  and 
when  motion  impends,  the  angle  between  dR   and  the  vertical  is  <t>  —  a. 
Taking  the  sum  of  the  vertical  components,  and  the  sum  of  the  moments 
as  above,  we  get  equations  which  yield  the  required  result. 

To  allow  for  the  friction  between  the  cap  and  the  head  of  the  screw,  let  n  = 
the  coefhcient  of  friction,  and  R  =  the  effective  arm  of  the  friction  there  with 
respect  to  the  axis  of  the  screw.     (If  the  surface  of  contact  between  the  cap 


82 


Chap,  iv 


and  the  head  were  flat  and  a  full  circle,  R  would  equal  two-thirds  the  radius 
of  the  circle.  But  the  contact  is  generally  a  hollow  circle,  as  in  Fig.  146,  and 
then  R  is  practically  equal  to  the  mean  radius.)     The  friction  moment  at  the 

cap  is  ixWR; 

(i)  for  raising  the  load,  Pa  =  Wr  tan  (0  +  a)  +  fiWR, 
(2)  for  lowering  the  load.  Pa  =  Wr  tan  (</>  -  a)  +  fiWR. 
§  4.  Journal  in  Worn  Bearing.  —  Fig.  147  represents,  in  section,  a. journal 
in  a  worn  bearing,  wear  much  exaggerated;  the  contact  between  the  two  is 
along  a  Hne  practically.  When  the  journal  is  about  to  turn  clockwise  and  slip, 
then  the  bearing  exerts  a  reaction  R',  making  an  angle  <^  (the  angle  of  fric- 
tion for  the  surfaces  in  contact)  with  the  normal  ON;  when  the  journal  is 
about  to  turn  counterclockwise  and  slip,  then  the  bearing  exerts  a  reaction 
R"  inclined  at  an  angle  (}>  with  ON,  but  on  the  other  side.  If  the  radius  of  the 
journal  is  r,  then  the  perpendicular  from  the  center  to  R'  and  R"  equals  r  sin  <^, 
and  the  circle  of  radius  r  sin  </>  with  center  at  the  center  of  the  cross  section  of 
the  journal  is  tangent  to  R'  and  R".  This  circle  is  called  the  friction  circle  for 
journal  and  bearing.  For  smooth  contacts  sin  (j)  nearly  equals  tan  (j>  or  /z,  and 
hence  the  radius  of  the  circle  practically  equals  iir. 


Fig.  147 


Fig.  148 


We  use  the  friction  circle  as  an  aid  to  fix  upon  the  line  of  action  of  the  re- 
action between  journal  and  bearing  when  motion  impends;  the  line  is  tangent 
to  the  circle.  For  example,  consider  the  bell  crank  shown  in  Fig.  148,  the 
journal  being  i|  inches  in  diameter  and  the  coefl&cient  of  friction  0.3;  the  re- 
quirement is  to  determine  the  least  force  P,  acting  as  shown,  which  will  over- 
come Q  (that  is,  start  the  bell  crank  to  turn  clockwise),  and  the  pressure  on  the 
bearing  then.  The  radius  of  the  friction  circle  is  f  sin  tan~^  0.3  =  0.18  inch. 
Since  there  are  but  three  forces  acting  on  the  bell  crank  {P,  Q,  and  R),  they 
are  concurrent,  that  is,  R  acts  through  0;  but  R  is  also  tangent  to  the  circle 
as  shown,  and  so  its  line  of  action  is  known.  To  determine  the  values  of  P 
and  R,  we  draw  AB  to  represent  Q  by  some  scale,  and  lines  through  A  and  B 
parallel  to  P  and  R  to  their  intersection  C;  then  BC  and  CA  represent  the 
magnitudes  and  directions  of  R  and  P  respectively, 

(Which  one  of  the  two  tangent  lines  to  take  can  be  determined  by  trial. 
Thus,  trying  ON,  the  contact  between  journal  and  bearing  would  be  at  N,  and 
the  tangential  or  frictional  component  of  the  pressure  on  the  journal  would 


Art.  20 


83 


be  as  shown,  not  consistent  with  the  assumed  tendency  to  slipping.  Obvi- 
ously the  other  tangent  is  the  correct  one,  and  on  investigating  for  the  friction 
component  of  R  when  acting  at  M  we  find  that  such  component  is  consistent 
with  the  assumed  tendency  to  slip.) 

The  force  P  which  would  just  permit  Q  to  start  the  bell  crank  to  turn  counter- 
clockwise could  be  determined  in  a  similar  way.  Then  R  would  act  along  the 
tangent  ON,  and  P  would  be  represented  by  C'A.  When  P  has  any  value 
between  C'A  and  CA,  then  slipping  does  not  impend,  and  the  line  of  action 
of  R  cuts  the  friction  circle. 

When  a  link  L  (Fig.  149)  of  a  machine  or  structure  is  pinned  to  other  parts 
or  members,  and  there  is  slipping  or  tendency  to  slipping  at  the  pins,  then  the 
pressure  exerted  by  each  pin  on  the  link  does  not  necessarily  act  through  the 
center  of  the  pinhole  there.  If  slipping  impends,  then  the  line  of  action  of 
the  pressure  is  tangent  to  the  friction  circle;  and  if  the  link  is  a  two-force 
member  (only  the  two  pin  pressures  acting  on  it),  then  the  two  pressures  are 
colinear  and  must  act  along  a  line  which  is  tangent  to  both  friction  circles. 
Which  one  of  the  four  tangents  to  take  in  a  given  case  depends  upon  the  direc- 
tion of  the  tendency  to  slipping  at  each  pin,  and  whether  the  link  is  under  ten- 
sion or  compression.  To  ascertain  the  correct  tangent,  try  any  one  as  the  line 
of  action  of  the  two  pin  pressures  R,  and  then  investigate  the  i?'s  for  their 
frictional  components  to  ascertain  whether  the  directions  of  those  components 
are  consistent  with  the  directions  of  slip;  only  one  tangent  will  satisfy  all 


Fig.  149 

the  conditions  for  a  given  case.  For  example,  suppose  that  the  tendency  is 
for  a  to  increase  and  /3  to  decrease;  if  the  pressures  put  the  link  under  tension, 
then  the  two  pressures  act  along  tangent  number  i  at  points  Ai  and  A2,  and 
if  the  pins  put  the  link  under  compression  then  the  two  pressures  act  along 
tangent  number  2  at  points  Bi  and  B^. 

The  deviations  of  the  various  tangents  (lines  of  action  of  the  pin  pressures) 
from  the  axis  of  the  link  depend  on  the  diameter  of  the  friction  circle  and  the 
length  of  the  link.  Generally  the  diameter  is  so  small  compared  to  the  length 
of  the  link  that  the  deviation  is  small,  and  one  may  safely  take  the  axis  of 
the  link  as  the  line  of  action  of  the  pin  pressures  so  long  as  the  link  is  at  rest 
and  for  all  states  of  tendency  to  slip. 

§  5.  Belt  or  Coil  Friction.  —  Fig.  150  represents  a  cyUnder  about  a  part 
of  which  a  belt  or  rope  is  wrapped.     If  the  cylinder  is  not  very  smooth,  then 


o .  Chap,  iv 

the  pulls  Pi  and  P2  may  be  quite  unequal  without  causing  slipping  over  the 
cylinder,  as  may  be  easily  verified  by  trial.  When  slipping  impends,  then  the 
ratio  of  these  pulls  depends  on  the  coefficient  of  friction  and  on  the  angle  of 
wrap.  If  P2  =  the  larger  pull,  m  =  the  coefficient  of  friction,  a  =  the  angle 
of  lap  expressed  in  radians,  and  e  =  base  of  the  Napierian  system  of  logarithms 
(2.718),  then  as  proved  below. 

For  a  given  value  of  Pi,  P2  increases  very  rapidly  with  a  as  shown  by  Fig. 
151,  which  is  the  polar  graph  of  the  foregoing  equation,  P2  and  a  being  the  vari- 
ables, e=  2.718,  M  taken  as  \,  and  Pi  =  OA.  The  following  table  gives  values 
of  the  ratio  P2/P1  for  three  values  of  the  coefficient  of  friction  and  for  twelve 
values  of  the  angle  of  lap. 

Maximum  Ratios  P2/P1  (Slipping  Impending) 


** 

M 

a 

a 

2  IT 

25r 

1 

1 

1 

1 

1 

4 

3 

2 

4 

3 

1 

O.I 

1. 17 

I  23 

1-37 

0.7 

3.00 

4-33 

9.00 

0.2 

1-37 

i-Si 

1.87 

0.8 

3SI 

5-34 

12.34 

0.3 

1.60 

1.87 

2.57 

0.9 

4. II 

6.58 

16.90 

0.4 

1.87 

2.31 

351 

I.O 

4.81 

8.12 

23  14 

o-S 

2. 19 

2.8s 

4.81 

2.0 

23- 

66. 

535- 

0.6 

2.57 

3-51 

6.59 

30 

III. 

535- 

12,390. 

Fig.  150 


Fig.  151 


Fig.  153 


Derivation  of  Formula.  —  The  forces  acting  upon  the  part  of  the  belt  in  con- 
tact with  the  cylinder  consist  of  the  tensions  Pi  and  P2,  the  normal  pressure, 
and  the  friction  (Fig.  152).  Let  p  denote  the  normal  pressure  per  unit  length 
of  arc;  then  the  normal  pressure  on  any  part  whose  length  is  ds  (enlarged  in 
Fig.  153)  is  pds.  The  friction  on  that  part  may  be  called  dF,  and  the  ten- 
sions P  and  P  +  dP.  Since  the  part  is  at  rest,  pds  =  2  P  sin  ^  dd  =  Pdd,  or 
p  =  P/r;  that  is,  the  normal  pressure  per  unit  length  at  any  point  of  the  con- 
tact equals  the  belt  tension  there  divided  by  the  radius  of  the  cylinder.  When 
slipping  impends,  dF  =  fxpds,  and  since  dF  =  dP, 

,„         P.  dP         ds 

dP  -=-  n  ~ds,  or   —-  =  ju  —  =  fjdd. 


Art.  20 


85 


Integration  gives  [ioge  Pj^'  =  fx  |^0j";  hence, 

loge  P2  -  log.  Pi  =  ^Jux,  or  P2  =  Pie*^. 

For  an  example  consider  the  band-brake  shown  in  Fig.  154.  It  consists  of 
a  rope  or  other  band  wrapped  part  way  around  a  brake  wheel  W,  the  two  ends 
of  the  band  being  fastened  to  the  brake  lever  L; 
the  lever  is  pivoted  at  Q.  Obviously  any  force  as 
P  tightens  the  band,  and  if  the  wheel  tends  to  turn 
(on  account  of  some  turning  force,  not  shown), 
then  P  induces  friction  between  wheel  and  band. 
We  will  now  show  how  great  a  frictional  moment 
(origin  in  the  axis  of  the  wheel)  the  force  P  can 
induce.  Let  M  =  the  moment,  P2  =  the  larger 
tension  in  the  brake  band  (on  the  side  as  marked 
when  the  wheel  tends  to  rotate  as  indicated),  Pi  =  the  smaller  tension,  r  = 
radius  of  the  wheel,  ai  =  arm  of  Pi  with  respect  to  Q,  (h=  arm  of  P2,  and 
a  =  arm  of  P.  Consideration  of  the  forces  acting  on  the  brake-strap  shows 
that  M  ={Pi  —  Pi)r;  consideration  of  forces  acting  on  the  lever  shows  that 
Pa  =  Pifli  +  P2a2.  For  a  given  P,  M  is  greatest  when  slipping  impends,  and 
then  P2  -V-  Pi  =  e*^.     These  three  equations  solved  simultaneously  show  that 

M  =  Pa  (e'*«  -  i)r  -=-  (a^e'"'  +  ai). 

For  example,  let  P  =  75  pounds,  a  =  10  feet,  n  =  I,  a  =  320°  (=  5.5 
radians),  r  =  3  feet,  ai  =  2  feet,  and  a2  =  9  inches.  Then  a  -5-  2  tt  =  about  9, 
and  e*"*  =  4.1 15  (see  table  on  preceding  page);  and 

M"  =  75  X  10  (4.1 1  —  i)  3  -T-  (I  X  4-11  +  2)  =  765  foot-pounds. 


Fig.  154 


CHAPTER   V 

CENTER  OF  GRAVITY 

21.   Center  of  Gravity  of  Bodies 

§  I.  It  is  shown  in  Art.  7  that  the  resultant  of  two  parallel  forces  Fi  and  Fj 
acting  at  two  points  A  and  B  of  any  body  cuts  the  line  A  Bin  a,  point  P  so  that 
AP/PB  =  F2/F1  (Fig.  155).  This  proportion  fixes  the  position  of  P,  and 
since  the  proportion  is  independent  of  the  angle  between  AB  and  the  forces,  P 
is  also,  so  independent.  Therefore  ii  AB  were  a  rod  and  Fi  and  F2  the  weights 
of  two  bodies  suspended  from  A  and  B,  then  the  resultant  R  of  Fi  and  F2  would 
always  pass  through  the  same  point  even  if  the  tilt  of  the  rod  were  changed 
slowly  so  as  to  leave  the  suspending  strings  parallel.  Furthermore,  if  three 
parallel  forces  be  applied  at  definite  points  A,  B,  and  C  of  a  body  (Fig.  155), 
and  if  R  denotes  the  resultant  of  Fi  and  F2  as  before  and  R'  the  resultant  of 
R  and  F3  (and  so  also  the  resultant  of  Fi,  Fo,  and  F3),  then  CP'/PP'  =  R/F3. 


This  proportion  fixes  P'  (in  CP),  and  it  is  independent  of  the  angle  between 
the  forces  and  the  plane  of  ABC.  Therefore  ii  AB  and  CP  be  two  rods  rigidly 
fastened  at  P,  and  Fi,  F2,  and  F3  the  weights  of  bodies  suspended  from  A 
B,  and  C,  then  the  resultant  of  the  three  forces  would  always  pass  through 
P'  if  the  rods  were  slowly  turned  about  leaving  the  strings  parallel.  And  so  if 
any  number  of  parallel  forces  have  definite  points  of  application  on  a  rigid 
body,  the  resultant  of  the  forces  always  passes  through  some  one  definite  point 
of  the  body,  or  of  its  extension,  when  the  body  is  turned  about  so  as  not  to 
disturb  the  parallelism  of  the  forces.  This  unique  point  is  called  the  center 
or  centroid  of  the  parallel  forces. 

The  forces  of  gravity  on  all  the  constituent  particles  of  a  body  constitute  a 
parallel  force  system  having  definite  points  of  application;  therefore  all  those 
forces  have  a  centroid.  That  is,  the  resultant  of  the  forces  of  gravity  on  all 
the  particles  of  a  body  (its  weight)  always  passes  through  some  one  definite 
point  of  the  body,  or  of  its  extension,  no  matter  how  the  body  is  turned  about; 

86 


Art.  21 


87 


this  point  is  called  the  center  of  gravity  of  the  body.  The  positions  of  the  cen- 
ters of  gravity  of  many  regular  bodies  are  given  in  Art.  24,  and  methods  for 
determining  those  centers  of  gravity  are  explained  in  Art.  23. 

We  now  show  how  to  locate  the  center  of  gravity  of  a  body  (or  of  a  collection 
of  bodies)  which  consists  of  simple  parts  whose  weights  and  centers  of  gravity 
are  known.  Let  yl,  B,  C,  etc.  (Fig.  156), 
be  the  centers  of  gravity  of  certain  parts 
of  a  body  (not  shown) ;  Wi,  W2,  Ws,  etc., 
the  weights  of  those  parts;  xi,  ji,  Zi,  the 
coordinates  of  A ;  X2,  }%  Z2,  the  coordi- 
nates of  B,  etc.  Also  let  W  denote  the 
weight  of  the  whole  body,  Q  its  center 
of  gravity,  and  x,  y,  z,  the  coordinates 
of  Q.  Since  W  is  the  resultant  of  Wi, 
W2,  W3,  etc.,  the  moment  of  W  about 
any  Hne  equals  the  algebraic  sum  of  the 

moments  of  Wi,  W2,  W3,  etc.,  about  the  same  line  (Art.  8). 
moments  about  the  y-axis,  we  get 


Fig.  156 


Thus,  taking 


Wx  =  Wixi  +  W2X2  +  W3X3  + 


•> 


from  which  equation  x  can  be  determined.  Similarly,  by  taking  moments 
about  the  x-axis  we  can  get  y.  To  get  z,  we  imagine  the  body  turned  until  the 
y-axis  is  vertical,  —  the  coordinate  axes  are  assumed  fixed  to  the  body,  —  and 
then  take  moments  about  the  x-axis;  or,  what  comes  to  the  same  thing,  we  im- 
agine the  forces  of  gravity  (TFi,  TF2,  TF3,  etc.)  all  turned  about  their  respective 
points  of  application  until  they  become  parallel  to  the  y-axis,  and  then  take 
moments  with  respect  to  the  x-axis. 

A  name  for  the  product  of  the  weight  of  the  body  and  the  ordinate  of  its 
center  of  gravity  with  respect  to  a  plane  will  prove  convenient;  we  will  call 
such  product  the  moment  of  the  body  with  respect  to  the  plane.*  Then  the 
equations  mentioned  can  be  rendered  in  the  form  of  a  proposition  as  follows: 
The  moment  of  a  body  with  respect  to  any  plane  equals  the  algebraic  sum  of 
the  moments  of  its  parts  with  respect  to  that  same  plane. 

(i)  As  an  example  we  determine  the  coordinates  of  the  center  of  gravity 
of  a  slender  wire  43  inches  long  bent  as  represented  by  the  heavy  line  in  Fig.  157. 
If  the  weight  of  the  wire  per  unit  length  is  w,  say,  then  the  weights  of  the 
several  straight  portions  beginning  at  the  left  are  as  listed  in  the  schedule 
under  W.  The  coordinates  of  the  respective  centers  of  gravity  are  listed  under 
X,  y,  and  2;  and  the  moments  of  the  parts  with  respect  to  the  yz,  zx,  and  xy 
planes  in  the  last  three  columns  respectively.     The  coordinates  of  the  center 

*  This  moment  does  not  of  course  have  anything  to  do  with  turning  effect  like  the  ordi- 
nary moment  of  a  force  (with  respect  to  a  line  or  point).  To  distinguish  these  moments,  the 
first  is  sometimes  called  a  statical  moment,  not  very  appropriately,  however.  See  also 
Art.  22  for  other  statical  moments. 


88 

of  gravity  of  the  whole  wire  are:  x=  177.5  ly  -^  43  «' 
43  z£)  =  3.44 in.;  z  =  192  2£;  -^  43  «'  =  4-47  in- 


Chap,  v 
4.13  in.;  y  =  148  w  -^ 


F 

X 

y 

z 

Wx 

Wy 

Wz 

6 

8 

10 

10 

4 

-2.5 
0.0 
0.0 

10.0 
10. 0 

0 

3 
6 
6 

3 

—  2 

8 
8 

4 
0 

4 
8 

—  12.5M; 
00.0 
00.0 
50.0 
100. 0 
40.0 

ow 
18 
48 
60 

30 
-8 

40  w 

48 

32 

00 
40 
32 

43  w 

177. Sw 

148  w 

192  w 

Fig.  158 


(ii)  As  another  example,  we  determine  the  center  of  gravity  of  a  flat  sheet 
of  tin  consisting  of  three  parts  (Fig.  158),  namely,  a  square,  a  semicircle,  and 
an  equilateral  triangle.  If  5  =  the  side  of  the  square,  and  w  =  the  weight  of 
the  tin  per  unit  area,  then  the  weights  of  the  parts  are  as  scheduled  under  W. 
Obviously,  the  center  of  gravity  of  the  square  is  at  the  intersection  of  the 
diagonals;  in  Art.  24  it  is  explained  that  the  center  of  gravity  of  a  semicircle 
is  4  r/3  TT  distant  from  the  center  of  the  circle  where  r  =  radius,  and  that  the 
center  of  gravity  of  a  triangle  is  on  the  medians  of  the  triangle  and  \  a  distant 
from  any  base  where  a  =  altitude  measured  to  that  base.  Hence  the  coor- 
dinates of  the  centers  of  gravity  of  the  parts  are  as  scheduled  under  x  and  y. 
The  moments  of  the  parts  with  respect  to  the  yz  and  zx  planes  are  scheduled 
in  the  last  two  columns  respectively;  hence,  for  the  whole  sheet  of  tin  x  = 
0.571  s^w  -^  1.826  shv  =  0.313  s,  and  y  =  0.633  ^"^  "="  1-826  shv  =  0.347  s. 


Part. 

W 

X 

y 

Wx 

Wy 

Square 

1 .  000  s^w 
•393 
•  433 

0.505 

•50 
-.289 

0.505 

—  .212 

.500 

.  500  s^w 
.196 
-.125 

.  500  S^li> 

Semicircle 

-•083 

Triangle 

.216 

i.826  5''w; 

.571 s^w 

.  633  s^w 

§2.  To  Determine  the  Center  of  Gravity  of  the  Remainder  of  a 
Simple  Body  from  which  one  or  more  simple  parts  have  been  taken,  we  may 
use  the  principles  of  moments  in  modified  form  thus:  The  moment  of  the  re- 


Art.  21 


89 


mainder  of  a  body  with  respect  to  any  plane  equals  the  moment  of  the  whole 
minus  the  moments  of  the  parts  taken  away. 

(iii)  As  an  example,  we  determine  the  center  of  gravity  of  a  cylinder  of  cast 
iron  (specific  weight  450  pounds  per  cubic  foot)  with  a  conical  recess  in  one 
end  and  a  cylindrical  hole  in  the  other,  shown  in  section     y 
in  Fig.  1 59.     The  weights  of  the  complete  solid  cylinder,    |/"]<-  4.'.'.>|<— -  5  "->\ 
of  the  cone,  and  of  the  small  cylinder,  all  as  of  cast  iron, 
are  given  under  W.     The  coordinates  of  the  center  of 
gravity  of  the  solid  cylinder  and  of  the  parts  are  given 
under  x  and  y  (see  Art.  24  for  information  on  cone),  and 
the  moments  with  respect  to  the  yz  and  zx  planes  are 
given  in  the  last   two  columns.     The  weight  of  the 
actual  piece  of  cast  iron  is  327.5  —  (41  +  26.2)  =  260.3 
pounds;  the    moments   of   the   piece  equal    1637.5  ~ 
(205.0  +  78,6)  =  1353.9  and  2620  -  (61.5  +  314.4)  = 
2244.1  inch  pounds  respectively.     For  the  piece  of  cast 
iron,    therefore,    x  =  1353.9  -^  260.3  —  5-2,   and   y  = 
2244.1  -r-  260.3  =  8.6  inches. 


Fig.  159 


Part. 

W 

* 

y 

Wx 

Wy 

Cylinder 

327-5 
41.0 
26.2 

5 
5 
3 

8 

i-S 
12 

1637-5 

205.0 

78.6 

2620.0 

Cone    

61. 5 
314-4 

Hole   

§  3.  Experimental  Methods  may  be  resorted  to  for  finding  the  center  of 
gravity  of  a  body  so  irregular  that  the  foregoing  methods  cannot  be  applied, 
(i)  Method  of  Suspension:  The  body  is  suspended  from  one  point  of  it, 
and  the  direction  of  the  suspending  cord  is  then  marked  in  some  way  on  the 
body;  the  operation  is  repeated  for  another  point  of  suspension.  Since  the 
center  of  gravity  is  in  the  two  lines  or  directions  so  fixed  in  the  body,  it  is  at 

their  intersection. 

(2)  Method  of  Balancing:  The  body  is 
balanced  on  a  straight-edge,  and  the  vertical 
plane  containing  the  edge  is  marked  on  the 
body;  the  operation  is  repeated  for  two  more 
balancing  positions  of  the  body.  Since  the 
center  of  gravity  is  in  the  three  planes  so  fixed 
in  the  body,  it  is  at  their  common  point. 
Fig.  160  This  method  is  readily  applied  to  a  body  in 

the  form  of  a  thin  plane  plate;  for  such  only  two  balancings  are  necessary. 

(3)  Method  of  Weighing:  The  weight  W  of  the  body  is  determined,  and 
then  it  is  supported  on  a  knife-edge  B  (Fig.  160)  and  on  a  point  support  which 
rests  upon  a  platform  scale;  the  reaction  W  of  the  point  support  is  weighed, 
and  the  horizontal  distance  a  of  the  point  from  the  knife-edge  is  measured; 


■  ''///////'//////n///, 


Chap,  v 
90 

then  the  horizontal  distance  from  the  center  of  gravity  to  the  knife-edge  is 
W'a/W.  In  this  manner  the  horizontal  distances  of  the  center  of  gravity 
from  several  knife-edge  supports  can  be  got  and  the  center  of  gravity  located. 
•  The  distance  of  the  center  of  gravity  of  a  body  from  the  plane  through  three 
points  of  the  body  can  be  determined  if  the  body  can  be  supported  at  the  points 
and  if  certain  weighings  can  be  performed  as  described.  Let  A,  B,  and  C 
(behind  B  and  not  shown)  be  three  such  points  of  the  body  (Fig.  161) ;  a  =  dis- 
tance of  A  from  the  line  joining  B  and  C;  W=  weight  of  the  body;  W'  = 
weight  recorded  by  the  scale  when  A,  B,  and  C  are  at  the  same  level  as  shown 


I 


Fig.  161 


Fig.  162 


in  Fig.  161,  and  W"  =  weight  recorded  by  the  scale  when  A  is  higher  than  B 
and  C  by  any  amount  h  (Fig.  162).  Then  the  distance  y  of  the  center  of  gravity 
from  the  plane  ABC  is  given  by 


Va2  -  h^  w'  -  W" 


a. 


'  h  W 

Proof:  From  the  first  position  it  is  plain  that  W'a  =  Wx;  from  the  second 
it  follows  that  W"a  cos  6  =  W{xcosd  -  ysinO).  Solving  these  simultane- 
ously we  get  y  =  {W  -  W")  (a  cote)/W;  but  cot^  =  Va-  -  h^  -^  h,  hence, 
etc. 

22.   Centroids  of  Lines,  Surfaces,  and  Solids 

§1.  Lines,  surfaces,  and  (geometric)  solids  have  no  weight,  and  therefore 
they  have  no  center  of  gravity  in  the  strict  sense  of  the  term  as  defined  in  the 
preceding  article.  However,  we  do  speak  of  the  center  of  gravity  of  those  geo- 
metric conceptions;  and  we  mean  by  the  term,  the  center  of  gravity  of  the  line, 
surface,  or  volume  materialized,  that  is,  conceived  as  a  homogeneous  slender 
wire,  thin  plate,  or  body,  respectively.  The  center  of  gravity  of  a  line,  surface, 
or  solid  is  sometimes  spoken  of  as  the  center  of  gravity  of  the  length  (of  the 
line),  area  (of  the  surface),  and  volume  (of  the  solid).  The  term  centroid  has 
been  proposed  as  a  substitute  for  center  of  gravity  when  applied  to  lines,  sur- 
faces, and  solids  as  being  more  appropriate;  the  new  term  is  given  preference 
in  this  book. 

If  a  given  line,  surface,  or  soHd  is  imagined  as  materialized,  then  we  can 
apply  the  principle  of  moments  (Art.  21)  to  it.  Thus,  \i  W  =  the  weight  of 
the  whole  materialized  line,  surface,  or  solid,  Wi,  Wi,  W2,  etc.,  =  the  weights 
of  all  the  parts  into  which  we  imagine  it  divided,  x  =  the  coordinate  of  the 


Art.  23  91 

center  of  gravity  of  the  whole  with  reference  to  some  convenient  reference 
plane,  and  Xi,  xi,  Xi,  etc.  =  the  coordinates  of  the  centers  of  gravity  of  all  the 
parts  respectively,  then 

W-x  =  PTiXi  +  Wtx^  +  Wzxz  +  •  •  •  . 

But  the  weights  W ,  W\,  W2,  W3,  etc.,  are  proportional  to  the  respective  lengths 
(L,  Li,  L2,  Lz,  etc.)  or  areas  {A,  Ai,  A2,  A3,  etc.)  or  volumes  (F,  Vi,  V2,  V3,  etc.), 
as  the  case  may  be ;  and  therefore  it  follows  from  the  preceding  equations  that 

for  lines,  Lx  =  LiXi  +  L2X2  +  L3X3  +  •  •  •  , 

for  surfaces,     Ax=  ^1X1  +  712X2  +  ^3X3+  •  •  •  , 

and  for  solids,  Vx  =  FiXi  +  F2X2  +  F3X3  +  •  •  •  . 

The  foregoing  formulas  can  be  rendered  conveniently  in  a  single  statement 
of  words  or  proposition  by  means  of  a  new  term  which  we  now  define.  The 
moment  of  a  line,  surface,  or  solid  with  respect  to  a  plane  is  the  product  of  the 
length  of  the  line,  area  of  the  surface,  or  volume  of  the  solid  and  the  coordinate 
of  the  centroid  of  the  line,  surface,  or  solid  with  respect  to  that  plane.  (The 
moment  of  a  plane  line  or  surface  with  respect  to  a  plane  perpendicular  to 
the  plane  of  the  line  or  surface  is  also  called  its  moment  with  respect  to  the 
line  of  intersection  of  the  two  planes.)  The  proposition  or  principle  of  mo- 
ments, then,  is  this:  The  moment  of  a  line,  surface,  or  solid  with  respect  to  any 
plane  equals  the  algebraic  sum  of  the  moments  of  the  parts  of  that  line,  surface, 
or  solid  into  which  we  imagine  the  whole  divided,  with  respect  to  that  same 
plane.*  The  principle  of  moments  can  be  used  to  determine  the  centroids  of 
all  geometrical  bodies  which  can  be  divided  up  into  parts  whose  magnitudes 
and  centroids  are  known.     Three  examples  follow: 

(i)  Let  it  be  required  to  locate  the  centroid  of  the  line 
represented  (heavily)  in  Fig.  163,  the  curved  portion  being 
a  circular  arc;  given  that  each  coordinate  of  the  centroid 
of  the  arc  is  6.366  inches  (Art.  24).  Let  x  denote  the  x 
coordinate  of  the  centroid  of  the  line;  then  taking  moments 
about  OY  (the  length  of  the  line  =  35.7  inches),  35.7  x  — 
10  X  o  +  10  X  5  +  15.71  X  6.366,  or  X  =  4.20  inches. 
Obviously,  the  y  coordinate  also  equals  4.20  inches.  ^^^'  ^^^ 

(ii)  Let  it  be  required  to  locate  the  centroid  of  the  shaded  area  in  Fig.  164, 
which  represents  the  cross  section  of  a  "channel"  (a  form  of  steel  beam  much 
used  in  construction).  We  consider  the  section  as  divided  into  a  rectangle, 
0.40  by  1 5  inches,  and  two  trapezoids.  The  distance  of  the  centroid  of  either 
trapezoid  from  its  longer  base  is  given  by  3  (0.90  +  0.80)  -^  3  (0.90  +  0.40)  = 
1. 3 1  inches  (Art.  24).  The  second  column  of  the  adjoining  schedule  gives  the 
areas  of  the  parts;  the  third,  the  centroidal  coordinates  with  respect  to  the  base 

*  Of  course  these  moments  have  nothing  to  do  with  turning  effects  like  the  moment  of  a 
force  with  respect  to  a  line  or  a  point.  To  distinguish  these  moments,  the  former  are  some- 
times called  statical  moments,  not  very  appropriately,  however. 


lO'-'-M 


92 


Chap,  v 


of  the  section;  and  the  last,  the  moments  with  respect  to  that  base.  The  dis- 
tance of  the  centroid  of  the  entire  section  from  the  base  is  7.70  -^  9.8  =  0.79 
inch. 


Part. 

A 

y 

Ay 

Rectangle       

6.0 
3-8 

0.20 
1. 71 

1 .20 
6.50 

Two  trapezoids 

9.8 

•       7-70 

Fig.  164 


k-^-->k /6" >k-,9'--->l 

Fig.  165 


(iii)  Let  it  be  required  to  locate  the  centroid  of  a  solid  consisting  of  a  cone, 
a  cylinder  and  a  hemisphere  as  represented  in  Fig.  165;  given  that  the 
centroid  of  the  cone  is  2  inches  from  its  base,  and  that  of  the  hemisphere  is 
3  inches  from  its  base  (Art.  24).  The  volumes  of  the  parts,  the  x  coordinates 
of  their  centroids,  and  the  moments  with  respect  to  the  yz  plane  are  as 
recorded  in  the  adjoining  schedule.  The  total  volume  is  4825.5  in^  and 
the  algebraic  sum  of  the  moments  of  the  parts  is  —6433  in*;  therefore  x  = 
—  6433  -^  4825.5  =  —  1.33  inches,  the  negative  sign  indicating  that  the  cen- 
troid of  the  whole  solid  is  to  the  left  of  0. 


Part 

V 

X 

M 

Cone 

S36.2 
3217.0 
1072.3 

10 

0 

—  II 

S.362 
0 

-11.795 

Cylinder 

Hemisphere 

4825. s 

-6,433 

Yk"  6'-'-->\<—  6" -H  Q  §  2.  To  Determine  the  Centroid  of  the  Remain- 

^  DER  OF  A  Simple  Figure  from  which  one  or  more 
simple  parts  have  been  taken,  we  may  use  the  prin- 
ciple of  moments  in  the  following  modified  form.  The 
moment  of  the  remainder  of  a  figure  with  respect  to 
any  plane  equals  the  moment  of  the  whole  minus  the 
moments  of  the  parts  taken  away.  For  example,  let  it 
Fig.  166  be  required  to  determine  the  centroid  of  the  shaded 

area  in  Fig.  i66,  the  part  of  the  square  remaining  after  the  triangle  and  the 


Art.  23 


93 


quadrant  have  been  taken  away;  given  that  the  centroid  of  the  triangle  is 
2  inches  from  OY  and  4  inches  from  YC,  and  that  the  centroid  of  the  quadrant 
is  2.54  inches  from  OX  and  CX  (see  Art.  24).  The  areas,  centroidal  coordi- 
nates, and  moments  appear  in  the  adjoining  schedule.  The  area  of  the  shaded 
portion  is  144  —  (36  +  28.27)  —  79-73  square  inches,  and  the  moments  of  the 
shaded  part  with  respect  to  the  y  and  x  axes  are  864  —  (72+266.9)  =  525.1 
and  864  —  (288+ 71.8)  =  494.2  cubic  inches  respectively.  Therefore  x  = 
525.1  -^  79.73  =  6.59,    and   y  =  494-2  -^  79-73  =  6.20  inches. 


Part 

A 

X 

y 

Ax 

Ay 

Souare 

144 

36 

28.27 

6 
2 

9-44 

6 
8 
2-54 

864 

-72 

—  266.9 

864 

Trianele 

-288 

Quadrant 

-71.8 

5251 

494.2 

23.   Centroids  Determined  by  Integration 

§  I.  If  it  is  desired  to  locate  the  centroid  of  a  line,  a  surface,  or  a  solid  which 
cannot  be  divided  into  a  finite  number  of  simple  parts  whose  lengths,  areas,  or 
volumes  and  centroids  are  known,  and  if  the  line,  surface,  or  solid  is  "mathe- 
matically regular,"  then  we  imagine  the  line,  surface,  or  solid  divided  into  an 
infinitely  great  number  of  parts,  and  apply  the  principle  of  moments.  To 
find  the  sum  of  the  moments  of  all  these  elementary  parts  involves  an  integra- 
tion. Thus,  let  L  =  the  length  of  a  line,  x  =  the  x  coordinate  of  its  centroid, 
dLi,  dLi,  dLz,  etc.  =  the  lengths  of  elementary  portions  of  the  line,  and  Xi, 
x-2,  xz,  etc,  =  the  x  coordinates  of  the  centroids  of  those  portions  respectively, 


then 


Lx  =  dLi  •  xi  +  dL2  •  X2  +  dLs  •  xz  -f 


=  / 


dL  •  X, 


in  which  dL  stands  for  any  of  the  elementary  lengths  and  x  for  the  x  coordinate 
of  the  centroid  of  that  dL.  Similarly,  for  areas  and  volumes;  and  thus  we 
have  these  formulas: 

(i)  Lx  =   jdL  .  x;    (2)  Ax=   j  dA  -  x;    (3)   Vx=   j  dV  -  x, 

and  corresponding  ones  for  y  and  i  (the  y  and  z  codrdinates  of  the  centroid). 

These  formulas  can  be  used  to  determine  x,  y,  and  z  if  the  form  of  the  line, 
surface,  or  solid  is  such  that  the  integrations  can  be  performed.  In  any  par- 
ticular case,  limits  of  integration  must  be  assigned  so  that  all  elementary  por- 
tions are  included  in  the  integration  (summation).*  Six  examples  illustrating 
their  use  follow: 

*  The  centroid  is  a  mean  point.  The  ordinate  from  any  plane  to  the  centroid  of  a  line, 
surface,  or  solid  equals  the  mean  of  the  ordinates  of  all  the  equal  elementary  portions  of  the 
line,  surface,,  or  solid,  it  being  understood  that  the  mean  takes  into  account  signs  of  the  ordi- 
nates.    For,  let  xi,  x^,  Xi,  etc.,  be  the  ordinates  of  the  elementary  portions  and  n  the  number 


94  ^^^^- '' 

(i)  Circular  arc;  radius  =  r  and  central  angle  =  2a  (Fig.  167).  The  radius 
which  bisects  the  central  angle  is  a  line  of  symmetry,  therefore  the  centroid  is 
on  that  line;  if  that  line  is  taken  as  x  axis,  then  y  =  o.  The  length  of  the  arc  = 
2  ra  (a  expressed  in  radians),  dL  =  rdcj),  andx  =  r  cos  (p;  therefore  formula  (i) 
becomes 


2  rax 


r d<f)'r  COS  (I)  =r^   j      cos  4>d(t)  =  2  r"^  sin  a;   or  x  =  (r  sin  a)  -i-  a. 

a  fJ  —a 

(ii)  The  preceding  problem  will  now  be  solved  without  using  polar  coordi- 
nates.    Since  x^  +  /  =  r\  xdx  +  ydy  =  o,  or  dy  =- (x dx)/y.     Hence 


and 


dL  =  Vdx^  +  ^y2  =  gxVi-^  xV/  =  dx  r/y  =  dx  rj^/r^  -  x\ 

wV    Cl'Jv 


2  rax 


=    I  xdL  =  2  r    / 

«y  *^  r  cos  a 


V  r^  —  x'^ 


=  2  r^sina;   etc. 


(iii)  The  parabolic  segment  AOBA  (Fig.  168);  altitude  =  a  and  base  =  b. 
Evidently  the  axis  of  the  parabola  is  a  line  of  symmetry,  and  therefore  it 
contains  the  centroid.  If  that  line  be  taken  as  the  x  axis,  then  y  =  o.  Let 
x  and  y  be  the  coordinates  of  any  point  P  on  the  parabola ;  then  the  area  of  the 
elementary  portion  shaded  is  2  y  dx.  Since  the  area  of  the  segment  is  f  ab, 
and  the  equation  of  the  parabola  is  4  ay^  =  b'^x,  formula  (2)  becomes 


-abx 


=    /    2y  dx'  X  =  —p.   \    xi  dx  —  -  ba?; 
Jo  -Va^o  5 


and 


2 

5 


—      ^  ^  'I 

X  =  -ba^  -r-  -ab  =  -a. 


2 
3 


of  them  (infinite);  then  the  mean  ordinate  is  {xi  -\-  X2  ~\-  x^  -{•  •  •  •)  -i-  n;  also,  let  Q  =  the 
length,  area,  or  volimie  of  the  line,  surface,  or  solid,  and  dQ  =  the  length,  area,  or  volume  of 
the  equal  elementary  portions;  then  the  mean  ordinate  equals 


fa +  3:2+  • 
ndQ 


•)dQ 


JdQ-x 


=    X. 


Art.  23 


95 


(iv)  Circular  sector  (Fig.  169);  radius  =  r  and  central  angle  =  2  a.  The 
radius  which  bisects  the  central  angle  is  evidently  a  line  of  symmetry,  and  so 
the  centroid  is  on  that  line.  If  that  line  is  taken  as  x  axis,  then  y  =  o.  The 
area  of  the  sector  equals  r^cc,  a.  expressed  in  radians;  dA  =  pd(f)'dp,  where 
p  =  OP  and  P  is  any  point  in  the  sector.     Therefore  formula  (2)  becomes 


and 


r^ax 


X 


I      pd(j)dp-  X  =    j      j      pd(j)dp' p cos <^; 

0    J— a  Jo    J— a 


=  ( -  /^  sin  a  j  -^  ( r^a  j  = 


2  r  sin  a 
3« 


(v)  Conical  or  pyramidal  solid;  altitude  =  a  (Fig.  170).  We  take  the  origin 
of  coordinates  at  the  apex,  and  the  x  axis  perpendicular  to  the  base;  OMNO 
represents  the  projection  of  the  cone  or  pyramid  on  the  XY  plane.  We  imagine 
the  sohd  divided  into  plates  or  laminas  parallel  to  the  base;  if  the  area  of  the 
base  is  called  ^,  say,  then  the  area  of  the  lamina  represented  is  Ax'^/a',  and  the 
volume  of  the  lamina  is  dx  •  Ax^/a^.  And  since  the  volume  of  the  solid  is  ^  ^a, 
formula  (3)  becomes 

T  r"  f*"  Aa^ 

-Aax  =    I    {dx'  Ax^/a^)  x  =  A/a^   /    x^dx  = ; 

3  Jo  Jo  4 


Fig.  169 


Fig.  171 


hence,  x  =  j  a,  that  is,  the  perpendicular  distance  from  the  centroid  to  the 
base  equals  one-fourth  the  altitude.  Evidently,  the  centroid  of  every  lamina 
lies  on  the  line  joining  the  apex  and  the  centroid  of  the  base;  therefore  the 
centroid  of  all  the  laminas  (that  is,  the  solid)  lies  on  that  line. 

(vi)  Octant  of  a  sphere;  radius  =  r  (Fig.  171).     Obviously  x  =  y  =  z;  ^  is 
given  by 

I  {dx  dy  dz)x. 

Jo 

Evaluating  the  integral  and  substituting  for  V  its  value,  \  irr^,  we  find  that 
x=  ^r. 

§  2.   Surfaces  and  Solids  of  Revolution.  —  For  surfaces,  we  use  formula 
(2)  and  select  as  element  the  surface  described  by  an  elementary  part  of  the 


96 


Chap,  v 


generating  curve  MN  (Fig,  172).     Let  the  x  axis  be  taken  coincident  with  the 
axis  of  revolution;   then  the  area  described  by  a  part  of  the  generating  curve 

of  length  ds  is  2  iry  ds.  The  centroid  of 
this  area  is  in  the  x  axis,  and  its  x  coor- 
dinate is  the  X  in  the  figure;  hence  if  A 
stands  for  the  area  of  the  surface  of 
revolution, 


I TT   I  yds-  X,  or  :*;  =  —r-  j  xy ds. 


Fig.  172 


Fig.  173 


The  limits  of  integration  must  be  assigned 
so  that  each  product  xy  ds  will  be  in- 
cluded in  the  integration. 

For  solids,  we  use  formula  (3),  and 
take  as  element  that  volume  generated 
by  an  elementary  part  of  the  generating 
plane  MPN  (Fig.  173)  which  is  included  between  two  lines  perpendicular  to 
the  axis  of  revolution.  Thus,  if  the  x  axis  is  taken  coincident  with  the  axis  of 
revolution,  then  PQqp  generates  the  elementary  volume,  or  dV  =  ir  {y^}  — 
y-^)  dx.  Now  the  centroid  of  this  elementary  volume  is  in  the  x  axis,  and  its 
X  coordinate  is  the  x  in  the  figure;  hence  if  V  denotes  the  volume  of  the  solid 
of  revolution,  then 

Vx  =  ir   I  (yi^  —  yi^)  dx'X,   or  x=yj  {yi^  —  yi^)x  dx. 

The  limits  of  integration  are  to  be  assigned  so  that  each  product  (yi^  —  yi^)x  dx 
will  be  included  in  the  integration.* 

*  Theorems  of  Pappus  and  Guldinus.  —  These  relate  primarily  to  the  determination  of 
the  area  and  volume  of  a  solid  of  revolution;  they  involve  the  centroid  of  the  generating  curve 
or  plane,  and  are  therefore  mentioned  in  this  place,  (i)  The  first  theorem  states  that  the 
area  of  a  surface  of  revolution  generated  by  a  plane  curve  revolved  about  a  line  in  its  plane 
equals  the  product  of  the  length  of  the  curve  and  the  circumference  of  the  circle  described  by 
the  centroid  of  the  curve.  Proof:  Let  MN  (Fig.  172)  be  the  generating  curve,  L  =  length 
of  the  curve,  y  =  the  ordinate  of  the  centroid  of  L  from  the  axis  of  revolution,  and  A  =  area 
of  the  surface  generated.     Then 

A  =   \  2-KydL  and  yL  =    \  y  dL. 

Combining  these  equations  we  get  yl  =  L  2  ivy,  which  is  the  proposition  in  mathematical 
form.  (2)  The  second  theorem  states  that  the  volume  of  a  solid  of  revolution  generated  by  a 
plane  figure  revolved  about  an  axis  in  the  plane  equals  the  product  of  the  area  of  the  figure  and 
the  circumference  of  the  circle  described  by  its  centroid.  Proof:  Let  MPN  (Fig.  173)  be  the 
generating  plane,  a  =  area  of  the  plane,  y  =  the  ordinate  of  the  centroid  of  a  from  the  axis 
of  revolution,  and  V  =  volume  of  the  solid  generated.     Then 

V  =   ( Tr{y2^  —  y)  dx,  and  from  eq.   (2),  a'y  =  j iji  —  yO  dx  h  (^2  +  yi). 

Combining  these  equations  we  get  F  =  o  2  Try,  which  is  the  proposition  in  mathematical  form. 

To  illustrate,  we  determine  the  area  of  the  surface  generated  by  revolving  the  circular  arc 

ABC  (Fig.  175)  about  AC,  and  the  volume  of  the  solid  generated  by  revolving  the  figure 


Art.  23  97 

For  an  illustration  imagine  the  quadrant  XY  (Fig.  174)  rotated  about  OX 
so  as  to  generate  a  hemisphere.  The  positions  of  the  centroids  of  the  surface 
and  solid  generated  could  be  computed  as  follows:  (i)  The  area  of  the  hemi- 
sphere is  2  irr^,  X  =  rsixKl),  y  =  r  cos  </>,  and  ds  =  r  d(t>;  hence  for  the  area 

IT 

^  =  (2  X  -^  2  TT/-^)    I  xyds  =  r  J    sin  ^  cos  (f>d<l)  =  ^r. 


Fig.  174  Fig.  175  Fig.  176 

(ii)  The  volume  of  the  hemisphere  is  f  xr',  j2  =  r  cos  <{),  x  =  r  sin  </>,  and  dx  = 
r  cos  (t>d(f);  hence  for  the  volume 


X  =  (tt  -4- 


f^rr^ 


)    /  (js^  —  o) ;» (/x  =  (3  r/2)  i    cos^  (f)  s\n<f)  d(f)  = 


§  3.  The  centroid  of  an  irregular  plane  surface  or  figure  can  be  determined 
graphically  or  experimentally.  The  graphical  method  requires  the  use  of  a 
planimeter  or  other  device  for  measuring  an  area.  Let  aaa'a'  (Fig.  176)  be 
the  figure  whose  centroid  is  to  be  located,  (i)  Take  OX  and  YX'  on  opposite 
sides  of  the  figure  at  any  convenient  distance  I  apart,  (ii)  Project  any  width 
of  the  figure  as  aa  on  YX' ;  connect  projections  hh  with  any  point  on  OX  as 
Q,  and  note  the  intersections  cc.  (iii)  Repeat  (ii)  for  other  widths  as  a'a' ^  and 
then  connect  all  points  like  c  by  a  smooth  curve,  (iv)  Measure  the  area  A ' 
included  by  this  curve,  and  the  area  A  of  the  given  figure.  Then  A' I  is  the 
(statical)  moment  of  A  with  respect  to  OX  (proof  follows),  and  hence  the  dis- 
tance from  OX  to  the  centroid  is  y  =  A'l  ^  A.  Proof:  Let  w  =  any  width 
of  the  figure  as  aa,  and  w'  the  corresponding  width  cc  of  the  derived  curve;  then 
the  moment  of  A  with  respect  to  OX  is 

y'wdy=    j    Iw' dy  =  I  j    w'dy  =  lA'. 

To  determine  the  centroid  experimentally  cut  a  piece  of  stiff  cardboard  into 
the  shape  of  the  irregular  figure,  and  find  its  center  of  gravity  by  balancing 
as  explained  in  Art.  2 1 ;   this  point  locates  the  centroid  sought. 

ABCA  about  AC.  The  length  of  the  arc  =  10.47  inches;  the  distance  of  its  centroid  from 
AC  =  0.89  inch  (Art.  24);  hence  A  =  10.47  X  2  tt  X  0.89  =  58.5  square  inches.  The  area  of 
ABCA  =  9.06  square  inches;  the  distance  of  its  centroid  from  AC  =  0.54  inch;  hence  V  = 
0.06  X  2  IT  X  0.54  =  30.7  cubic  inches. 

The  theorems  A  =  L  2i^y  and  F  =  a  2  ttj  can  be  used  also  for  computing  y  if  A  and  Z,, 
or  V  and  a  (as  the  case  may  be),  are  known. 


98 


Chap,  v 


24.   Centroids  of  Some  Lines,  Surfaces,  and  Solids 

Circular  Arc  (Fig.  177).  —  C  is  the  centroid;  its  distance  from  the  center  is 
{r  sin  a) /a,  the  divisor  a  to  be  expressed  in  radians  (i  degree  =  0.0175  radian); 
.^JA  the  distance  is  also  rc/s,  where  5  =  arc.     If  the  arc  is  flat 

then  the  distance  of  its  centroid  from  the  chord  is  nearly 
f  h;  the  discrepancy  is  less  than  one-half  per  cent  for 
arcs  whose  central  angle  2  a  is  less  than  60  degrees. 

When  the  arc  is  a  semicircle,  then  the  distance  from 
the  centroid  to  the  center  is  2  r/x  =  0.6366  r.     When  the 
arc  is  a  quadrant,   then   the   distance  to  the  center  is 
2  r  V'2/x  =  0.9003  r,  and  the  distance  to  the  radii  OA  and 
OB  is  2  r/ir  =  0.6366  r. 
Triangle.  —  The  centroid  is  at  the  intersection  of  the  medians;    its  per- 
pendicular distance  from  any  side  equals  one-third  the  altitude  of  the  triangle 
measured  from  that  side.     If  rci,  x^,  and  x^  are  the  coordinates  of  the  vertexes 
with  respect  to  any  plane  and  x  the  coordinate  of  the  centroid,  then  x  = 

\  {Xx-\-  X2  +  -T3). 


Fig.  177 


Trapezoid.  —  The  centroid  is  on  the  median  (line  joining  the  middle  points 
of  the  parallel  sides)  (Fig.  178); 


1  = 


(B-  h)a 
6(5  +  6)' 


m  = 


{2B  +  h)a 


n  = 


(5+  2h)a 
siB  +  b)  ' 


Two  geometrical  constructions  for  locating  position  on  the  median  follow: 
(i)  Extend  AE  (Fig.  179)  so  that  BE  =  CD,  and  in  the  opposite  direction 
extend  CD  so  that  DF  =  AB;  the  intersection  of  FE  and  the  median  GH  is 
the  centroid  sought.  (2)  Divide  the  trapezoid  (Fig.  180)  into  triangles  by  a 
diagonal  as  ^C;  find  the  centroids  d  and  G2  of  the  triangles  (construction 
indicated  in  the  figure) ;  the  intersection  G1G2  with  the  median  EF  is  the  cen- 
troid sought. 

Quadrilateral.  —  (i)  Divide  the  quadrilateral  into  triangles  by  a  diagonal  AC 
(Fig.  181)  and  find  their  centroids  d  and  G2;  divide  it  into  triangles  by  the 
other  diagonal  and  find  their  centroids  Gz  and  d;  the  intersections  of  the  lines 
G1G2  and  G3G4  is  the  centroid  sought.  (2)  Divide  the  sides  into  thirds  (Fig. 
182),  and  draw  lines  through  the  third  points  as  shown;  these  lines  form  a 
parallelogram  whose  diagonals  intersect  at  the  centroid  of  the  quadrilateral. 


Art.  24 


99 


Sector  of  a  Circle  (Fig.  183).  —  C  is  the  centroid;  its  distance  from  the  center 
is  I  {r  sin  a) /a,  the  divisor  a  to  be  expressed  in  radians  (i  degree  =  0.0175 
radian) ;  the  distance  also  equals  f  rc/s,  where  s  =  arc. 

A 


Fig.  181 


When  the  sector  is  a  quadrant,  then  the  distance  of  the  centroid  from  the 
center  is  4  V2  r/3  r  =  0.6002  r;  and  the  distance  to  the  radii  OA  and  OB  is 
4  ;-/3  TT  =  0.4242  r.  For  a  semicircle  the  distance  from  diameter  to  centroid 
is  4^/3x  =  0.4242  r. 


Fig.  183 


Fig.  185 


Sector  of  a  Circular  Ring  (Fig.  184).  —  The  distance  from  the  centroid  to  the 
center  is 


2  R^ 


r'^  sm  a 


a 


3   i?2  _  y2 

the  divisor  a  to  be  expressed  in  radians  (i  degree  =  0.0175  radian). 

Segment  of  a  Circle  (Fig.   185).  —  The  distance  from  the  centroid  to  the 
center  is 


(?         2  f '  sin^  a 


12  A  3^ 

where  A  denotes  the  area  of  the  segment.    ^  =  |  r^  (2  a  —  sin  2  a),  the  first 
a  to  be  expressed  in  radians  (i  degree  =  0.0175  radian). 


Ok— I  a  — •» 


I 

lA 


Fig.  187 


lOO 


Chap,  v 


The  Area  Shaded  in  Fig.  i86,  included  between  a  quadrant  and  the  tangents 
at  its  extremities.  The  distance  of  the  centroid  from  the  bounding  tangents 
is  0.223  r,  and  the  distance  to  their  intersection  is  0.315  r. 

Parabolic  Segments  (Fig.  187).  —  Ci  and  d  are  the  centroids  of  the  shaded 
parts.  Their  distances  from  the  sides  of  the  inclosing  rectangle  (a  X  b)  are 
marked  in  the  figure. 

Elliptic  Segment  (Fig.  188).  —  The  centroid  of  the  segment  XAAX  coincides 
with  the  centroid  of  the  segment  XaaX  of  the. circumscribed  circle;  the  cen- 
troid of  the  segment  YBBY  coincides  with  the  centroid  of  the  YbbY  of  the 
inscribed  circle. 


Fig.  li 


Fig.  189 


Right  Circular  Cylinder  (Fig.  189).  —  C  is  the  centroid;  its  distance  from 
the  axis  of  the  cylinder  is  \  {r-  tan  a)/h,  and  its  distance  from  the  base  is 
1  ^  _|- 1  (^2  ^^^2  Q,)//^  When  the  oblique  top  cuts  the  base  in  a  diameter  of 
the  base  (lower  part  of  Fig.  189),  then  the  distance  from  the  centroid  to  the 
axis  is  x\  irr,  and  to  the  base  30  ^^• 

Cone  and  Pyramid.  —  The  centroid  of  the  surface  (not  including  base)  is  on 
a  line  joining  the  apex  with  the  centroid  of  the  perimeter  of  the  base  at  a  dis- 
tance of  two-thirds  the  length  of  that  line  from  the  apex.  The  centroid  of  the 
solid  cone  or  pyramid  is  on  the  lines  joining  the  apex  with  centroid  of  the  base 
at  a  distance  of  three-fourths  the  length  of  that  line  from  the  apex. 

Frustum  of  a  Circular  Cone.  —  Let  R  =  radius  larger  base,  r  =  radius  smaller, 
a  =  altitude.  The  distance  of  the  centroid  of  the  curved  surface  from  larger 
base  is  I  aiR  +  2  r)/(R -\- r);  from  smaller  base  I  a{2  R  +  r)/iR  +  r); 
from  a  plane  midway  between  bases  i  a{R  -  r)/{R  +  r).  The  distance  from 
the  centroid  of  the  solid  frustum  to  the  larger  base  is 

I  a(R'  -\-2Rr-\-s  r^)/{R^  +  Rr  +  r^). 

Frustum  of  a  Pyramid.  —  If  the  frustum  has  regular  bases,  let  R  and  r  be 
the  lengths  of  sides  of  the  larger  and  smaller  bases,  and  h  the  altitude;  then 
the  distance  of  the  centroid  of  the  surface  (not  including  bases)  from  the  larger 
base  is  \h{R-\-  2  r)/(R  -\-  r).  Whether  the  bases  are  regular  or  not,  let  A 
and  a  =  the  areas  of  the  large  and  small  bases  and  h  the  altitude;  then  the 
distance  of  the  centroid  of  the  solid  from  the  larger  base  is 

lh{A-\-2  VJa  +  3  a)/(A  -f  Vl^-f  a). 


Art.  24 

Obelisk  and  Wedge  (Fig.  190).  —  The  distance  from  ilie 
centroid  to  the  base  AB  is 

^h(AB-hAb-\-aB-}-3  ab)/(2  AB -\- Ab  + aB -\- 2  ab). 

li  b  =  o  the  solid  is  a  wedge,  and  the  distance  from  the 
centroid  to  the  base  is 

^h(A-\-a)/(2A-ha). 


•loi 


Fig.  190 


Sphere.  —  The  centroid  of  any  zone  (surface)  of  a  sphere 
(Fig.  191)  is  midway  between  the  bases.  The  distance  of  the  centroid  of  a 
segment  (solid)  (Fig.  192)  from  the  base  is  I  h{^  r  —  h)/(;^  r  —  h)\  when  h  =  r 
(hemisphere)  then  the  distance  is  f  r.  The  distance  of  the  centroid  of  a  sector 
(solid)  (Fig.  193)  from  the  center  of  the  sphere  is  f  (i  +  cos  a)  r  =  f  (2r  —  h). 


Fig.  192 


Fig.  193 


Ellipsoid.  —  Let  the  three  axes  be  taken  as  x,  y,  and  z  coordinate  axes,  and 
a,  b,  and  c  to  denote  the  semi-lengths  of  the  corresponding  axes  of  the  ellipsoid; 
the  centroid  of  one  octant  of  the  solid  is  given  by  x  =  f  ^>  3'  ~  f  ^>  3-^^  2  =  f  c. 

Paraboloid  of  Revolution,  formed  by  revolving  a  parabola  about  its  axis. 
Let  h  =  height  of  the  paraboloid,  the  distance  from  its  apex  to  the  base,  then 
the  distance  from  the  centroid  of  the  solid  to  the  base  is  |  /f. 


CHAPTER   VI 

SUSPENDED   CABLES   (WIRE,   CHAIN,  ETC.) 
25.   Parabolic   Cable 

§1.  Symmetrical  Case.  —  When  a  cable  is  suspended  from  two  points  and 
it  sustains  loads  uniformly  spaced  along  the  horizontal  and  spaced  so  closely 
that  the  loading  is  practically  continuous,  then  the  curve  assumed  by  the 
cable  is  a  parabolic  arc  as  will  now  be  shown.  The  symmetrical  case  (points 
of  suspension  at  same  level)  will  be  considered  first.     Let  AOB  (Fig.  194)  be 


Y 

,H 

—  X 

'        ( 

D 

> 

'WX 

X 

Fig.  igs 


the  cable  suspended  from  A  and  B,  w  =  load  per  unit  (horizontal)  length, 
a  =  span  AB,f  =  sag,  H  =  tension  in  cable  at  lowest  point,  and  T  =  tension 
at  any  other  point  Q  (coordinates  x  and  y).  The  forces  acting  on  the  portion 
OQ  are  H,T,  and  the  distributed  load  wx  (Fig.  195);  this  load  acts  at  mid- 
length  of  X.  Since  the  forces  are  in  equilibrium,  their  moment-sum  equals 
zero  for  any  origin  of  moments;  hence  moments  about  Q  give  Hy  =  wx{x/2), 


,      2H 

X-  =  ■ — ■  y, 

w 


or 


y 


•w 


(i) 


This  is  the  standard  form  of  the  equation  of  a  parabola;  the  axis  of  the  parab- 
ola coincides  with  the  y  axis,  and  the  vertex  is  at  O.  If  we  substitute  for 
X  and  y  their  values  for  the  point  A  {x  =  a/2,  and  y  =  f),  then  we  get 
a2/4  =  (2  H/w)  f,ox  H  =  wa^/Sf;  hence  equation  (i)  may  be  written 


x^ 


a^ 
4/ 


}': 


or 


4/, 
a2  • 


(2) 


A  formula  for  the  tension  T  at  any  point  Q  may  be  arrived  at  as  follows: 
Let  (}>  =  slope  of  the  curve  at  Q;  then  it  is  plain  from  Fig.  195  that 
r  sin  (^  =  wx,  and  T  cos  <^  =  H.     Squaring  and  adding  gives 


102 


Art.  25  ^°3 

At  the  points  of  suspension,  x  =  a/ 2,  and  the  value  of  T  at  that  point  is 

£?  ( I  +  16  ^;V  =  *  wa  (i  +  -^J-  (4) 


The  adjoining  table  gives  values  of  T/wa  for  various  values  of  f/a,  the  sag 
ratio  (denoted  by  n  in  the  table). 


n  = 
T/wa  = 


I.O 

0-5 

0.25 

0.  125 

0. 1 

0.05 

0.515 

0.559 

0.707 

I.  ir8 

1.346 

2.550 

O.OI 
12. 81 


The  length  of  cable  for  any  span  a  and  sag  /or  sag  ratio  n  =  f/a.  — Let  /  = 
length  of  cable  AB  and  ds  =  length  of  an  elementary  portion;  then  as  in  all 
plane  curves,  ds^  =  dx^  +  dy^  =  [i  +  (dy/dx)-]  dx^,  or 

ds  =  [1  +  (dy/dxY]^  dx. 

From  the  equation  of  the  curve  (2),  dy/dx  =  (Sf/a^)  x  =  S  nx/a;  hence 

ds  =  (i  +  64nH^/a'^)^  dx. 

Integrating  between  proper  limits  (o  and  |  /  for  s,  and  o  and  |  a  for  x)  and  then 
doubling,  we  get 

I  =  a(i(i  +  i6n^)'+^\ogeUn-\-{i  +  16  n^)'])-  (5) 

An  approximate  formula  for  I,  much  more  convenient  to  use  than  the  fore- 
going one,  m^ay  be  deduced  as  follows:  Expanding  the  coeflScient  of  dx  above 
by  the  binomial  theorem,  we  get 

(/5  =  fi +32^2— 2— 5i2w^-4+   •  •  •  jdx; 


I 


^< 


i-\--n^-^n^  + 


)■ 


(6) 


and  integrating  between  limits  as  before  we  find  that 

8    „      3^ 

3  5 

The  following  table  gives  values  of  l/a  by  the  exact  and  approximate  formulas, 
equations  (5)  and  (6)  respectively,  for  several  sag  ratios  n  =  f/a. 


1. 01 
I .0003 
1. 0003 


§2.  Unsymmetrical  Case.  —  By  this  is  meant  a  cable  suspended  from 
two  points  not  at  the  same  level;  see  Fig.  196,  where  ACB  represents  a  cable 
suspended  from  A  and  B.  In  this  case,  also,  the  cable  hangs  in  the  arc  of  a 
parabola  as  will  be  proved  presently.  Let  a  =  horizontal  distance  between 
points  of  supports  (as  in  §1),  b  =  vertical  distance  between  the  points,  6  = 
angle  which  AB  makes  with  horizontal  (  =  tan~^  b/a),  x  and  y  =  coordi- 
nates of  any  point  Q  of  the  cable  as  shown,   T  =  tension  at  the  highest 


n  = 

1.0 

0.5 

0.25 

0.125 

0. 1 

0.05 

l/a  exact 

2.3234 

1.4789 

1.1478 

I . 0402 

1.0260 

I . 0066 

l/a  approximate 

. 

1.1417 

I. 0401 

1.0260 

I . 0066 

I04 


Chap,  vi 


point,  V  and  H'  respectively  =  the  two  components  of  T  along  AY  and 
AB.     There  are  three  forces  acting  on   the  part   AQ,  -  its  load  wx,  the 


tension  T,  and  the  tension  at  Q.     The  moment-sum  for  these  three  forces  for 

any  origin  equals  zero;   with  Q  as  origin 

-  wx  •  x/2  +  V'x  -  H'  (QP)  =  o,  or  V'x  -  wx^/2  =  H'  (y  cos  ^  -  x  sin  6).    (i) 

This  is  the  equation  of  a  parabola  with  the  axis  parallel  to  the  y  coordinate 
axis. 

To  express  the  equation  of  the  curve  in  terms  of  the  dimensions  a,  b,  and 
the  vertical  sag  /i  under  the  middle  point  of  the  chord  AB:  —  The  forces 
acting  on  the  entire  cable  consist  of  the  load  wa,  the  tension  at  A,  and  that  at 
B.     Their  moment-sum  with  origin  at  B  is 

wa-a/2  —  V'a  =  o;       hence      V^  =  •wa/2.  (2) 

The  forces  on  the  upper  half  AC  consist  of  the  load  wa/2,  the  tension  at  A, 
and  that  at  C.     Their  moment-sum  with  origin  at  C  is 

72 


wa'' 


'^'•£_rf  +  H'/,cos9  =  o;       hence      ff' =  g^— -,- 


Substituting  these  values  of  V  and  H'  in  (i)  gives 


^^  {a  —  x)  =  y  —  re  tan  d, 
a^ 


,-2 


or        >•  =  (4/1  +  ^)- -4/1-2 


(3) 


(4) 


The  vertical  distance  of  any  point  as  Q  below  the  chord  AB  \s  y  Xxtand; 
hence  if  we  let  y'  denote  that  distance,  the  foregoing  equation  can  be  put  into 
the  more  convenient  form 

4/1^ 


y  = 


a' 


(a  —  x). 


(5) 


Art.  25 


105 


The  value  of  the  slope  at  any  point  of  the  curve  is  (differentiating  equa- 
tion 4) 


dy  _    /i 


8/1  JC 


_  =4--  -^ ^-. 

ax        a       a        a  a 

Let  a  and  /3  =  the  slope  angles  at  .4  and  B  respectively  (where  x  =  o,  and 
X  =  a);  then 

tan  a  =  {b-\-  4/i)/a,       and       tan  6  =  (b  —  4/i)/a.  (6) 

Let  Xo  and  jo  =  the  coordinates  of  the  vertex  of  the  parabola  (where 
dy/dx  =  o);   then 

Xo  =  a{b-\-  4/i)/8/i,       and       yo  =  (b  +  4/i)-/i6/i.  (7) 

Let  H  and  V  respectively  =  the  horizontal  and  vertical  components  of  T. 
Then  (see  Fig.  196) 

H  =  H'cosO  =  wa'^/Sfi,  and 

V  =  V  -\-  H'  sine  =  ^  wa  {i  -\-  a  tan  0/4/1); 

and  since  T^  =  H"^  -\-  V^,  we  find  that 

sin  0        11 
wr       2  wi         J 

where  «i  =  sag  ratio  fi-i-  AB  = /i  -^  asec0.  The  last  expression  shows 
that  for  given  w,  a,  and  Wi,  the  tension  T  increases  as  the  angle  d  is  made 
larger;  also  that  for  given  w,  a,  and  6,  T  increases  as  Wi  is  made  smaller.* 

Length  of  the  Parabolic  Arc  AB  (Fig.  196).  —  Let  ai  =  the  length  of  the 
chord  AB,  Wi  =  sag  ratio  /i  -r-  ci,  and  h  =  length  of  the  arc  AB.  Also  let 
ds  =  length  of  an  elementary  portion  of  the  arc;  then 

(/s  =  [i  +  {dy/dx)^f  dx. 
From  the  equation  of  the  curve  (4),  we  can  get 

This  last  value  of  dy/dx  substituted  in  the  foregoing  expression  for  ds  gives 
ds-  ^ 


^-wi'-'h'^'ilh'A^n 


} 


i  +  8wi(i  —  2-J2  «i/ 1  —  2  -  j  +  sin  0   > 


sec  6  dx. 


Now  this   equation  is  in  the   form  ds  =  {i-\-  X)^secd  dx,  where 

X  =  S  til  (1  —  2  x/a)  [2  «i  (i  —  2  x/a)  +  sin  6]. 

*  Let  MN  (Fig.  197)  represent  the  load  on  the  cable  ^B,and  let 
MO  and  NO  be  parallel  to  the  tangents  at  A  and  B  (Fig.  196)  re- 
spectively; then  OMNO  is  a  force  triangle  for  the  three  forces  act- 
ing on  the  cable  AB,  and  OM  represents  T  and  NO  represents  S. 
It  is  plain  from  the  figure  that  OR  tan  a  —  OR  tan  0  =  MN,  or 
OR  (tan  a  -  tan  /S)  =  MN.  But  OR  =  T  cos  a  =  S  cos  /3,  and 
MN  =  wa;  hence 

T  cos  a  (tan  a  —  tan  0)  =  wa  =  S  cos  /3  (tan  a  —  tan  18). 
Substituting  the  values  of  tan  a.  and  tan/3  given  by  equation  (6),  we 

find  that 

r  cos  a  =  wa^l&Ji.  =  5  cos  0.  FiG.  197 


io6 


Chap,  vi 


Unless  the  sag  is  relatively  large  ds  and  sec  d  dx  are  nearly  equal  at  all  points 
along  the  curve  (see  Fig.  196);  hence  (i  +  X)  is  nearly  equal  to  i  at  all  points, 
which  means  that  X  is  small  compared  to  i.  Therefore,  we  may  expand 
(i  _j_  X)  by  the  binomial  theorem,  and  drop  all  terms  except  the  first  few 
without  serious  error.  Thus  we  have  as  a  close  approximation  ds  =  {i -\- 
iX-^X^)secddx, 


and 


'■  =  / 


(i  +  |Z-  lX^)secddx. 


Substituting  for  X  and  X^  their  values,  and  integrating  we  finally  get 

/i  =  fli(i  +  §cos2  0.«i2-  V-«i'). 


(10) 


If  the  approximation  made  in  the  derivation  of  formula  (10)  is  not  per- 
missible in  a  given  case,  then  one  might  determine  the  exact  length  of  the 
cable  AB  somewhat  as  follows  when  a,  b,  and/i  are  given:  We  first  locate  the 
vertex  O  of  the  parabola  of  which  the  cable  is  a  part  from  equation  (7).  The 
vertex  will  be  found  either  between  A  and  B,  on   the  cable  (Fig.  198),  or 


A'v 


B'"> 


Fig.  198 


Fig.  199 


—Tfi^^ 


beyond  B  (Fig.  199).     Then  we  determine  the  length  of  the  arcs  AOA'  and 
BOB'  by  means  of  equation  (5),  §1,  and  finally  the  length  h  of  the  arcAB  from 

Zi  =  i  AOA'  +  h  BOB'  for  Fig.  198,  or  h  =  ^  AOA'  -  \  BOB'  for  Fig.  199. 

For  example  take  a  =  800  feet,  h  =  300  feet,  and  /i  =  200  feet.     Let  Xq  and 
jQ  =  the  coordinates  of  the  vertex.     From  equation  (7) 

(soo  +  4  X  200)  800  ,  (300  +  4X200)2 

Xf,  =  -^^^ =  tji^o,  and  Vo  =  — 7-7^ =  i1°-S- 

°  8X200  ^^  16X200 

Hence  the  cable  hangs  as  shown  in  Fig.  200.     The  length  AA'  =  1348.6  feet 
according  to  (5),  (a  =  iioo  and  n  =  378.5  -4-  iioo);   the  length  BB'  =  530.9 


■^...250'->k- 5S0' — ->< 

Fig.  200 


feet    according    to    (5),    (a  =  500    and    n  =  78.5  -f-  500). 
i  X  1348.6  +  i  X  530.9  =  939.8  feet. 


Hence    AB  = 


Art.  26 


107 


26.     Catenary  Cable. 

§1.  Symmetrical  Case.  —  A  chain  or  flexible  cable  suspended  from  two 
points  and  hanging  freely  under  its  own  weight  or  a  load  uniformly  distributed 
along  its  length  assumes  a  curve  called  (common) 
catenary.  Let  A  and  B  (Fig.  201)  be  the  points  of 
suspension  of  such  a  cable,  C  its  lowest  point,  Q  any 
other  point  of  the  cable,  5  =  the  length  CQ,  H  =  ten- 
sion at  C,  T  =  tension  at  (),  0  =  slope  of  the  curve 
a.t  Q,  w  =  weight  of  load  per  unit  length  of  cable,  and 
c  =  a.  length  so  that  cw  =  H  or  c  =  H/w.  The  forces 
acting  on  CQ  are  H,  T,  and  ws.  Since  they  are  in 
equilibrium,  T  cos  cj)  —  H,  and  T  sin  (f)  =  ws;  hence 
tan  0  —  ws/H  =  s/c.      But  tan  <^  =  dy/dx,   therefore  Fig.  201 

dy/dx  =  s/c.  (i) 

Now   since   ds^  =  dx"^  +  dy^,  {ds/dyY  =  (dx/dyY  +  i    and    {ds/dxY  =  i  + 
{dy/dxY;  also 


ds^  _  c^  _c^  -\-  s"^ 

dyl        S'  s'^ 


and 


ds 
dx 


c^-\-s^ 


(2) 


Integrating  the  first  one  of  these  equations  we  get  y  =  (c^  -\-  5^)2  -{-  A  where 
^  is  a  constant  of  integration.  But  y  =  c  where  s  =  o,  therefore  A  =  o, 
and  hence 

>»-  =  c^  +  S-,         or         s-  =  y2  _  ^2,  (3) 


or         s-  =  y 

Integrating  the  second  differential  equation  we  get 

s 


X 


cloge 


V 


/■ 


+  1 


=  c  sinh~^  - 


the  constant  of  integration  being  zero  (x  =  o  when  s  =  o).     From  (3) 


X 


s  =  Ic  {e'l'^  —  e  ^1'^)  —  c sinh - 


(4) 


(5) 


To  obtain  the  cartesian  equation  of  the  catenary  we  combine  (3)  and  (4) 
or  (3)  and  (5)  so  as  to  eliminate  5.  Thus  squaring  (5)  and  comparing  with 
(3)  we  get  easily 

(6) 


X 


y  =  \c  ie^''^  -\-  e  ^^'^)  =  ccosh-. 


or 


.  =  clog,[^±\/(^)'-i]  =  .cosh-f. 


(7) 


The  slope  angle  4>  at  any  point  in  terms  of  the  coordinates  of  the  point 
{x,y,s)  is  given  by 

^an<^  =  s/c  =  \  if'"  -  e-""^')  =  sinh  {x/c).  (8) 


xo8  Chap,  vi 

See  equations  (i)  and  (5).     And  from  equations  (2)  and  (3),  we  get 

sin0  =  s/y    and     cos0  =  c/y.  (9) 

It  follows  from  the  equilibrium  equation  T  sin  <^  =  ws  and  (9) ,  that 

T  —  wy,  (10) 

that  is,  the  tension  at  any  point  Q  equals  the  weight  of  a  length  of  cable  reach- 
ing from  Q  to  the  directrix  OX.  Hence  T  increases  from  C  to  A.  According 
to  the  definition  of  c 

H  =  wc.  (11) 

In  passing,  it  may  be  noted  that  since  T  cos  ^  =  H,  the  horizontal  com- 
ponent of  the  tension  at  any  point  Q  =  wc,  constant 
^  for  a  given  suspended  cable. 

As  in  the  preceding  article,  let  a  =  span  AB  (Fig. 

202),  /  =  sag,  and  /  =  length  of  cable  ACB.  Any 
two  of  the  three  dimensions  a,  I,  and  /  determine  the 
catenary,  as  will  be  shown  presently.  For  the  point 
A,  X  =  I  a,  y  =  f -{- c,  and  s  =  ^  I.  Hence  substi- 
tuting in  equations  (3),  (4),  and  (6)  respectively  we 

get 

Fig.  202 

(/  +  ,)2  =  ,2  +  i;2^  or  clj=\{lljy-\.  (3') 

\a  =  c sinh-i  (1  //^)^      or  1  ^Ic  -  sinh-i  {\ l/c).  (4') 

and  f-\-c  =  c  cosh  (^  a/c),        or     i  +  (//c)  =  cosh  (|  a/c).  (6') 

When  I  and  /  are  given  (3')  gives  c,  and  then  a  may  be  gotten  from  (4')  or 
(6').  When  a  and  /  are  given  (6')  determines  c  but  the  equation  cannot  be 
solved  directly,  —  only  by  trial  or  by  some  sunilar  method;  having  thus 
determined  c,  I  may  be  gotten  from  (3')  or  (4').  When  a  and  /  are  given, 
(4')  determines  c  (solution  by  trial),  and  then  /  may  be  gotten  from  (3')  or 

(6'). 
Inasmuch  as  these  trial  methods  are  generally  long,  computations  on  some 

catenary  problems  may  be  facilitated  by  means  of  diagrams.  In  Fig.  203 
the  curves  marked  A  give  the  relation  between  J/ a  and  l/a  for  values  of 
f/a  from  o  to  0.5  and  (corresponding)  values  of  l/a  from  i  to  about  1.50. 
For  example,  let  a  =  800  feet  and  /  =  160  feet.  Then  f/a  =  0.20,  and  the 
corresponding  ordinate  (over  f/a  =  0.20)  to  curve  A  reads  i.io;  hence 
l/a  =  I.IO,  and  /  =  800  X  i.io  =  8S0  feet  (length  of  cable). 

Most  practical  catenary  problems  involve  the  strength  of  the  wire  or  cable 
and  the  load  per  unit  length  of  wire.  For  such  problems  we  have,  in  ad- 
dition to  (3'),  (4')  and  (6'), 

T  =  w(f+c),     or     T/w=f+c,  (11') 

where  T  =  the  greatest  tension  (at  the  points  of  support),  which  should  of 


Art.  26 


109 


course  not  exceed  the  strength  of  the  wire.  Most  of  these  problems  can  be 
solved  by  trial  only,  unless  a  diagram  is  available.  For  example,  given  the 
strength  T  of  a  wire,  the  load  per  unit  length  w,  and  the  span  a;  required 
the  proper  length  of  wire  /.     Here 

T/wa  =  f/a-\-  c/a.  -  (11") 

This  equation  and  (6')  contain  only  two  unknown  quantities  /  and  c,  and  the 
two  equations  determine  /  and  c.  But  they  can  be  solved  only  by  trial. 
After/  and  c  have  been  ascertained,  then  I  may  be  computed  from  (3')  directly. 
The  curves  marked  B  in  Fig.  203  show  the  relation  between  f/a  and  T/wa. 

I.-30 


MS 


1.20  O 

•1- 


wo 

I.1>0 

1.00 

~ 

A 

. 

1 

J. 

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\ — 

1 

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- 

1 

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K 

■ 

1 

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1.0 

5. 

1 

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■  9 

5 

l:^b 

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^l.it> 

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/ 

f 

^, 

— 

" 

t- 

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uf; 

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i.ih 

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, 



"v 

^ 

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— 

^ 

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— 

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1.15 


1.10 


O 
> 


.05 


1.00 


.05  .10  .15  .20  .25  .30  .35  .40  .45 

Values  of  -f  -^  a 
Fig.  203 

Thus  ii  f/a  =  0.20,  as  in  the  preceding  illustration,  then  the  corresponding 
ordinate  to  curve  B  (over //a  =  0.20)  reads  0.85;  hence  T/wa  =  0.85  and 
T  =  0.85  wa*. 

*  Fig.  203  was  prepared  from  plate  II  of  Mr.  Thomas'  paper  mentioned  in  the  footnote 
at  the  end  of  this  chapter.  (For  cases  of  relatively  small  sag  ratios,  see  that  plate.)  The 
curves  marked  A  might  be  prepared  as  follows:  (i)  Assume  different  values  of  ///;  (ii)  com- 
pute the  corresponding  values  of  c/f  by  means  of  (3');  (iii)  compute  values  of  l/c 
IromJ/c  =  (///)  -i-  (c//);  (iv)  compute  values  of  a/c  from  (4');  (v)  compute  values  of  l/a 
from'  l/a  =  Q/c)  -^  {a/c);  (vi)  compute  values  of  f/a  from  J/ a  =  (l/a)  -r-  (///).  Finally 
plot  values  of  f/a  and  l/a.  (The  adjoining  schedule  gives  the  computed  values  for  ///  =  5 
as  an  illustration.) 


1 

n 

HI 

IV 

v 

VI 

VU 

vni 

l/f 

c/f 

l/c 

a/c 

l/a 

f/a 

c/a 

T/wa 

5 

2.6250 

1.9048 

1.6946 

1-1237 

0.2247 

0.5900 

0.8147 

To  plot  curves  B  we  would  continue  the  computation,  first  computing  values  of  c/a  from  the 
values  of  a/c  already  obtained,  and  then  values  of  T/ic-a  from  T/im  =  f/a  +  c/a.  I'inally 
values  in  columns  vi  and  viii  of  the  schedule  when  plotted  furnish  the  desired  graph. 


no 


Chap,  vi 


§  2.  Unsymmetrical  Case  (points  of  suspension  not  on  same  level).  — The 
cable  uniformly  loaded  along  its  length  hangs  in  an  arc  of  a  catenary.  The 
vertex  C  may  be  on  the  cable  (between  the  points  of  suspension  A  and  B)  as 
in  Fig.  204,  or  beyond  the  lower  point  of  suspension  as  in  Fig.  205.     In  either 


r 
\ 
\ 


'^' 


o 


-(l-k)a  ■ 


.-..■rHik-Oa 


ka 


Fig.  204 


Fig.  205 


figure,  a  =  the  horizontal  distance  between  A  and  B,b  =  the  vertical  distance, 
6  =  angle  between  AB  and  the  horizontal,  /  =  sag  or  vertical  distance  AC, 
and  I  =  arc  AB.  Most  problems  in  this  case  as  in  the  symmetrical  case 
can  be  solved  only  by  a  trial  method;  hence  diagrams  are  practically  neces- 
sary in  this  case  also. 


3.00 


"2.50 


I  2.00 


1.50 


:2  1.00 


50 


r       j;;    -t^          r          Tr    "    "IJIT          rZ"    ~       J 2" 

t        «        i                                         i^          4^             -iT             ^ 

I       V        1       '      1           ^         V^^         r  ^         ,   ^        r             J 

b^a\0    X05    \jg       .15          .20         i.25 '         UO  '         *35         /40        45/ 

jptr              ^      1        ^        ^        2> 

X    '    \     t                     t        '       7        ^y 

t        *        J           4                                              t                '                 t                 W- 

5     ^     it     J                             -         J         -t           ^^' 

^^     u    C     ^     ^-                 -        I        t      ^^'  -  ^ 

K   ^   \     ^     \      L       _      t      t   yi^  y 

\     ^     ^       ^      \        >                   tT  ^^t  ^% 

\      t      S        V       V        +           V          t        X>^   ^   y   ^ 

\        ^        \           ^           \           \              \               L_        ^^,'      %^'^^ 

\         A          ^             ^           \             N-              \               \      ^-^     V%^5%'^^ 

^     \:  ^^     ^       5       ^         3^      V"^^^-^,-^  ^^^•^ 

X     ^^  ^^     ^       S       ^         S     ^%^ ^^\^^Z2^Z,^^ 

\     5     "^^     5     \        ^^      \^^^^^  y^^'l.^'XL 

!s     ^^^s     ^^  ^«.     ^  ^^;;J^^-%^<t--^^ 

"^s     ^     ^^     ^        ^^     y'^^<^'^>i-^^'^-^'^ 

===========""'^^:-^:-=:p^^^-^^^^^i:^^i^^ 

~p:0^'^''^'.!>^O*:iM^^'^.'^1»-  .J--'  ■V''     i" 

bi^   S* "''''' '^''^''■' ^'''    ^"^      J^ 

^  ^O^""    ^.J^   _,^j'     ^''_^.'  '    |j'   ^1              i- 

.^•a^ -  ,,^ '^ - t' -• "]  !•' ' '"         —  \' 

--  =  -s^f=g^-^-§^^^^-r--r ^ --  =  " 

1.60 


.50 


1.40 


1.30 


1.20  > 


— -  1. 10 


.05  .10  .15  .20  .25  .30 

Values  of  f  -i-  a 
Fig.  2o6 


.35 


.40 


.45 


.50 


1.00 


In  Fig.  2o6  there  are  two  groups  of  curves  relating  to  this  unsymmetrical 
case;  the  group  occupying  the  lower  right-hand  portion  consists  of  graphs 
showing  the  relation  between  //a  and  Z/o  (values  at  right-hand  margin)  for 
ten  values  of  hi  a  (slope  of  AB,  Figs.  204  and  205).     The  other  group  consists 


Art.  26 


III 


of  graphs  showing  the  relation  between  //a  and  Tjwa  (values  at  left-hand 
margin)  for  the  same  ten  slopes  (r  =  tension  at  higher  point  of  support  and 
w  =  weight  of  cable  per  unit  length).  To  illustrate,  let  a  =  200  feet,  6  =  40 
feet,  /  =  240  feet,  and  w  =  2  pounds  per  foot.  On  the  curve  for  b/a  =  0.20 
in  the  lower  group,  we  find  the  point  whose  ordinate  l/a  =  1.20  and  note 
that  the  abscissa  of  that  point  is  f/a  =  0.385.  Hence  /  =  200  X  0.385  =  77 
feet.  On  the  curve  (or  b/a  =  0.20  of  the  upper  group,  we  find  the  point 
whose  abscissa  is  0.385  and  note  that  its  ordinate  T/wa  =  0.90.  Hence 
T"  =  2  X  200  X  0.90  =  360  pounds.* 

*  Fig.  206  was  made  from  certain  of  the  (more  extensive)  figures  in  Mr.  Robertson's  paper 
mentioned  in  the  footnote  at  the  end  of  this  chapter.  The  following  is  an  explanation  of 
a  method  for  the  construction  of  such  a  figure.  Let  h  =  arc  AC  and  h  =  arc  BC  (Figs  204. 
and  205);  also  let  yi  and  y-i  =  ordinates  of  A  and  B  respectively,  and  ka  and  (t  —  k)a  —  the 
abscissas  of  A  and  B.  Then  for  .4,  x  =  ka,  y  =  yi  and  s  =  h;  for  B,  x  —  (k  —  i)a,  y  =  ^2, 
and  s  =  h.     Hence,  substituting  in  equations  (3)  and  (6)  of  §r,  we  get 


(^M?r 


and 


2!i 
c 


=  cosh  k  -  ' 
c 


and 


—  =  cosh  (k  —  i)  -  ■ 
c  c 


(3') 
(6') 


At  the  higher  point  of  support  A  (Fig.  204  or 
(10)  §1,  the  tension  at  that  point  =  wyi,  and 


205),  y  =  yi',  hence  according  to  equation 


tca       a       \c  j      \c  j 


This  equation,  (3')  and  (6'),  constitute  the  basis  of  the  method.  We  first  assume  a  value 
of  k,  say  0.6,  and  different  values  of  a/c  (say  0.02,  0.04,  etc.);  then  (i)  compute  values  of 
yjc  and  y2/c  from  (6')  corresponding  to  those  values  of  a/c  (and  k  =  0.6);  (ii)  compute 
the  values  of  h/c  and  h/c  from  (4')  to  correspond;  (iii)  compute  values  of  l/c  from 
l/c  =  (/i/c)  +  (k/c);  (iv)  compute  values  of  f/c  iromf/c  =  {yjc)  -  i.  Finally  compute-  b/a 
from  b/a  =  [(yi/c)  -  (^2^)]  -^  {a/c)  (see  Figs.  204  and  205);  l/a  from  l/a  =  {l/c)  ^{a/c)\ 
f/a  ;from  f/a  =  {f/c)  -^  {a/c);    and   T/wa  from   T/wa  =  {yi/c)  ^  {a/c).      (See  schedule) 


1 

a/c 


u 

yi/c 


Ul 

yi/c 


IV 

h/c 


V 

h/c 


VI 

l/c 


vn 

f/c 


VUl 

b/a 


IX 

l/a 


X 

f/a 


XI 

T/wa 


0.6 

+ 
-D  0.4 

% 
S  0.3 


^^1 


0. 


^' 

/ 

^y 

0^ 

y 

/ 

)^ 

/ 

r 
/ 

y 

/) 

/ 

f 

1.1  1.3  1.5 

Values  of  1-s-a 

Fig.  207 


J 

V 

y 

-^ 

/ 

^. 

y 

A 

^ 

y 

^ 

y 

O.l  0.3  0.5 

Values  of  f-5-a 

Fig.  20S 


—r— 
/ 

/ 

f 
/ 

\ 

( 

\ 

\. 

H 

r^ 

V 

^ 

M 

'^ 

"■■■ 

■*>_■ 

1.0  1.2  1.4 

Values  of  T*wa 
Fig.  209 


0.6 
0.5 
0.4 
0.3 
0.2 
0.1 


We  now  plot  values  of  b/a  and  l/a  as  in  Fig.  207,  and  get  a  curve  (marked  k  =  0.6);  plot 
values  of  h/a  and  f/a  as  in  Fig.  208,  and  get  a  curve  (marked  k  =  0.6),  plot  values  of  b/a 
and  T/wa  as  in  Fig.  209,  and  get  a  curve  (marked  k  =  0.6).     In  a  similar  manner  we  take 


112 


Chap,  vi 


§3.  Approximate  Solutions  of  Catenary  Problems.  —  If  the  cable  is 
suspended  from  two  points  at  the  same  level  and  the  sag  is  small  compared 
to  the  span  so  that  the  slope  of  the  catenary  is  small  at  every  point,  then  the 
load  (weight)  per  unit  length  of  span  is  nearly  constant  and  equal  to  the 
weight  of  the  cable  per  unit  length.  Hence  the  catenary  coincides  very 
nearly  with  a  parabola  of  the  given  span  and  sag,  and  the  formulas  and  re- 
sults of  the  preceding  article  §1  (symmetrical  case)  may  be  applied  to  the 
case  here  under  consideration  without  serious  error. 

That  the  catenary  agrees  closely  with  a  parabola  can  be  shown  otherwise 
as  follows:  Expanding  the  exponentials  in  the  equation  of  the  catenary, 
(6)  §1,  we  get 


pX/c 


— +  — + 
20^      3  c^ 


and 


X 


I  ^    ,     x^         oc^     , 

C  2C^         3  c"* 


hence  the  equation  of  the  catenary  may  be  written 


3'  =  7    2+^  + 


X 


Neglecting   the  higher  powers  of   the   small   quantity   -,  we  have  as   close 

approximations 

y  =  c  -\-  x'^/2  c,      or      x"^  =  2  c  (y  —  c). 

These  are  equations  of  a  parabola  whose  axis  coincides  with  the  y  coordinate 
axis  and  vertex  c  distant  above  the  origin  of  coordinates. 

If  the  supports  A  and  B  are  not  at  the  same  level  (Fig.  ig6)  and  the  sag  / 
of  the  cable  is  small  compared  to  the  distance  between  the  points  of  support, 
then  the  slope  of  the  catenary  is  nearly  constant  and  the  load  per  unit  length 
of  horizontal  distance  is  nearly  constant  {w  sec  6,  where  iv  =  weight  of  cable 
per  unit  length,  and  0  =  angle  BAX).     Hence  the  catenary  coincides  very 

k  =  0.7  say,  and  make  computations  i,  ii,  iii,  etc.,  as  described;   then  plot  three  more  curves 
(Figs.  207,  208  and  209).     Then  we  repeat  for  still  other  values  of  k. 

From  the  three  sets  of  curves  (Figs.  207,  208,  and  209)  we  pick  out  sets  of  corresponding 
values  of  l/a,  f/a  and  T/wa  for  the  several  values  of  b/a.  Thus  for  b/a  =  0.2,  we  find  the 
adjoining  tabulated  values  from  the  curves. 


b/a  ■- 

=  0.20 

k 

Ua 

f/a 

T/wa 

0.6 
0.7 

etc. 

I    IS 
1.06 
etc. 

0.35 
0.25 
etc. 

0.91 
1.30 
etc. 

Then  we  plot  these  values  of //a  and  l/a  and  get  the  curve  marked  b/a  =  0.20  in  lower  group, 
Fig.  206;  also  we  plot  values  oi  f/a  and  T/wa  and  get  the  curve  marked  b/a  =  0.20  in  the 
upper  group  of  Fig.  206.  In  a  similar  way  we  tabulate  and  plot  for  other  values  of  b/a  (0.30, 
0.40,  etc.)  and  complete  the  diagram  (Fig.  206). 


Art.  27 


113 


nearly  with  a  parabolic  arc  of  the  given  (oblique)  chord  AB  and  sag  /i,  and 
the  formulas  of  the  preceding  article  §2  (unsymmetrical  case)  may  be  ap- 
plied to  the  cable  under  consideration  without  serious  error,  it  being  under- 
stood that  w  of  §2  =  (weight  of  cable  per  unit  length)  X  sec  6. 

27.     Cables  with  Concentrated  Loads 

§1.  Weight  of  Cable  Negligible.  —  Let  Fig.  210  represent  a  cable  ACB 
suspended  from  two  given  pomts  A  and  B,  C  being  a  given  point  from  which 
a  load  is  suspended.  If  the  cable  can  be  "  laid  out"  in  a  drawing,  the  ten- 
sions m  ylC  and  BC  can  be  determined  easily  by  constructing  the  force  tri- 
angle PQRF  for  the  load  W  and  the  two  tensions.  PQ  =  W  according  to 
some  convenient  scale;  PR  and  QR  (parallel  to  ^C  and  BC  respectively) 
represent  the  tensions  in  ^C  and  BC.  Or,  if  one  wishes  to  avoid  graphical 
methods  the  two  tensions  {Ti  and  T2)  may  be  computed  by  solving  the  tri- 
angle algebraically.     Such  solution  would  give 

Ti  =  W  cos  i3/sin  (a  +  |3)       and       T2  =  W  cos  a/sin  (a  +  /3), 
where  a  and  /3  are  the  angles  which  AC  and  BC  make  with  the  horizontal 
(Fig.  210). 


Fig.  210 


Fig.  211 


When  several  bodies  are  suspended  from  given  points  on  the  cable,  the 
cable  takes  up  a  definite  position,  but  it  is  not  easy  to  determine  the  slopes  of 
the  segments  of  the  cable  and  the  tensions.  The  difficulty  lies  in  the  alge- 
braic computation.  For  example,  consider  the  case  represented  in  Fig.  211. 
The  given  data  are  shown  in  the  figure;  the  lengths  are  drawn  to  scale,  but 
the  inclinations  of  the  segments  of  the  cable  may  not  be  correct,  being  un- 
known as  yet.  Let  the  inclinations  be  called  a,  13,  and  7  as  shown;  and  Ti,  T2, 
and  7^3  =  the  tensions  in  OA,  AB,  and  BN  respectively.  At  each  point  of 
suspension  of  a  load  {A  or  B)  there  are  three  forces  acting;  at  A,  the  load 
1000  pounds,  Ti,  and  T^,  and  at  B,  the  load  2000  pounds,  T2,  and  ^3.  Con- 
sideration of  forces  at  A  and  of  those  at  B  gives  respectively 

Ti  cos  a^  To  cos  i3        and         Ti  sin  a  -  Tz  sin  /3  =  1000 
Ti  cos  /3  =  Ti  cos  7       and         T2  sin  /3  -h  73  sin  7  =  2000. 


114 


Chap,  vi 


It  is  plain  from  the  geometry  of  the  figure  that 
8  cos  a  +  lo  cos  /3  +  1 2  cos  7  =  20,      and      8  sin  a  +  10  sin  j8  —  12  sin  7  =  4. 

These  six  equations  may  be  solved  simultaneously  for  the  six  unknowns 
(Ti,  T2,  Ts,  a,  jS,  and  7);  the  actual  solution  is  not  simple.  For  similar  cases 
with  more  than  two  loads,  the  work  of  solving  the  equations  increases  rapidly 
with  increasing  number  of  loads. 

Suspended  loads  can  be  chosen  so  as  to  hold  points  of  suspension  {A,  B, 
etc.)  in  certain  definite  positions.  For  instance  let  it  be  required  to  de- 
termine Wi,  W2,  etc.,  to  hold  a  cable  in  the  position  shown  in  Fig.  212.     We 


Fig.  212 


may  assume  any  value  for  one  of  the  weights  and  then  compute  the  values 
of  the  others.  Thus  taking  Wi  =  1000  pounds  say,  then  we  compute  the 
tension  in  AB  from  a  force  triangle  for  the  three  forces  acting  at  A.  PQXP 
is  such  a  triangle,  where  PQ  =  1000  pounds  (according  to  any  convenient 
scale)  and  PX  and  QX  are  parallel  to  OA  and  AB  respectively;  then  XQ 
represents  the  tension  in  AB.  The  next  step  is  to  find  the  value  of  W2  which 
corresponds  to  such  tension  in  AB;  so  we  draw  a  force  triangle  for  the  three 
forces  acting  at  B  one  of  which  is  the  determined  tension  in  AB.  This  force 
triangle  is  QXRQ,  and  so  RQ  represents  W2  and  XR  represents  the  tension  in 
BC.     Finally,  we  draw  the  force  triangle  RXSR  for  the  three  forces  acting 

at  C,  one  of  which  is  the  determined 
tension  in  BC,  and  thus  find  that  W3 
is  represented  by  SR.  Obviously  any 
three  weights  Wi,  W2,  and  TVs  in  the 
proportion  of  PQ,  QR,  and  RS  would 
hold  the  cable  in  the  specified  position. 
§2.  Weight  of  C.A3le  Not  Neg- 
ligible. —  It  is  assumed  in  the  fol- 
lowing discussion  that  the  cable  segments  are  quite  flat  so  that  they  are 
practically  parabolic  arcs  (see  preceding  article  §3).  Then  the  weight  of  any 
segment  of  the  cable  is  practically  the  same  as  the  weight  of  a  length  equal  to 
the  chord  of  the  segment.  Let  ABC  (Fig.  213)  be  a  cable  supported  at  A  and 
C,  a  load  being  suspended  from  the  cable  at  its  middle  point  B.  Given  the 
span  AC  =  2  a,  the  length  of  the  cable  =2!,  the  weight  of  the  cable  per  unit 


Art.  27  115 

length  =  w,  and  the  load  =  W;  required  the  sag  (depth  of  B  below  AC)  and 
the  tension  at  A.  This  (apparently)  simple  problem  is  determinate  but  prac- 
tically unsolvable  on  account  of  algebraic  difficulties.  The  equations  are 
easily  set  up.  Thus  let  ai=  the  (unknown)  length  of  chord  AB,fi  =  the  sag 
of  the  cable  below  the  chord  as  in  Fig.  196,  S  =  the  tension  and  ^  =  the  slope 
of  the  cable  at  B  (Fig.  196).  Then  according  to  equations  (10)  and  (6)  re- 
spectively of  Art.  25,  §  2 

According  to  the  footnote,  on  page  105. 

5cosi3  =  (7£'ai/a)aV8/i-  (3) 

From  the  three  forces  acting  at  B  (IF,  S,  and  S),  it  is  plain  that 

2Ssin^  =  W.  (4) 

These  four  equations  determine  the  unknowns  appearing  in  them,  d,  /i,  S,  and 
/3.  Thus  by  division,  the  last  two  give  tan  jS  =  4  Wfi/waia;  equating  the  two 
values  of  tan  /S  and  transforming,  we  get 

»iZ/.=v/;rM'_,/..  (5) 

ai  wa  ai       ▼  \ai/  ai 

This  equation  and  (i)  contain  only  two  unknowns,  the  ratios  (a/ai)  and  (/"i/ai), 
and  the  equations  determine  the  ratios.  Supposing  the  ratios  determined  we 
may  find  ai  since  a  is  given,  and  then  /i.  Exact  simultaneous  solution  of 
equations  (i)  and  (5)  is  impossible,  but  each  equation  may  be  graphed  and 
then  the  coordinates  of  their  intersection  would  be  the  desired  values  of  c/ci 

and/i/ai. 

The  converse  of  the  preceding  problem  is  much  simpler.  It  is  this:  Given 
the  span  AC  =  2  a,  the  chord  AB  =  ax,  the  sag/i,  and  the  weight  of  the  cable 
per  unit  length  w;  required  the  load  W.  Equations  (2),  (3),  and  (4)  give  in 
succession  (3,  S,  and  W.     Equation  (i)  gives  the  length  /.* 

*  For  other  information  on  the  subjects  of  this  chapter,  particularly  as  related  to  elec- 
tric transmission  lines,  see  the  following:  University  of  Illinois  Bulletin,  No.  11  (191 2)  by 
A.  Gruell;  Transactions  American  Institute  of  Electric  Engineers,  Vol.  30  (191 1),  papers  by 
Wm.  L.  Robertson,  Percy  H.  Thomas,  and  Harold  Pender  and  H.  F.  Thompson.  These 
papers  contain  extensive  tables  and  diagrams,  and  discuss  effects  of  temperature  changes. 


DYNAMICS 


DYNAMICS 


CHAPTER   VII 

RECTILINEAR    MOTION 
28.   Velocity  and  Acceleration 

§  I.  Velocity. — When  a  point  moves  so  that  it  traverses  or  describes 
equal  distances  in  all  equal  intervals  of  time  then  it  is  said  to  move  uniformly, 
and  we  call  the  motion  uniform.  All  other  motions  we  call  nonuniform.  By 
velocity  of  a  moving  point  is  meant  the  time-rate  at  which  the  point  is  moving, 
or  describing  distance.  To  express  the  magnitude  of  any  velocity  we  must  of 
course  compare  that  velocity  to  some  particular  velocity  as  a  standard  or  unit. 
Any  velocity  —  that  of  light,  for  example  —  might  be  taken  as  standard;  but 
it  is  more  convenient  to  take  the  velocity  of  a  point  moving  uniformly  and 
describing  a  unit  of  length  in  a  unit  of  time  for  a  standard.  Thus,  we  use  the 
foot  per  second,  mile  per  hour,  etc.  The  word  per  in  these  names  of  velocity- 
units  is  quite  commonly  replaced  by  the  soHdus  sign  /;  thus,  foot  per  second, 
mile  per  hour,  etc.,  are  abbreviated  to  ft/sec,  mi/hr,  etc.* 

In  any  uniform  motion  the  velocity  may  be  computed  by  dividing  the  dis- 
tance traversed  in  any  interval  of  time  by  the  interval.  Thus,  if  i;  =  the 
velocity,  A5  =  the  distance  traversed,  and  A/  =  the  interval  of  time,  then 

V  =  As /At.  (i) 

In  any  nonuniform  motion  the  rate  of  moving  is  not  constant  but  changes 
continuously,  as  we  all  realize.  Not  all,  however,  have  a  clear  notion  of  the 
value  of  the  rate,  or  velocity,  at  a  particular  instant  of  the  motion.  To  bring 
this  matter  up  definitely,  let  us  consider  the  following  example:  —  In  a  certain 
launching,  the  ship  moved  through  the  distances  given  after  5  in  the  adjoin- 
ing schedule  in  the  times  given  after  t. 

/  =  o       2        4  6  8         10  12  14         16  seconds; 

5  =  0     3.4      9.3      17.3      27.4      39.6        53.4         69.4      88.0  feet. 

Any  displacement  divided  by  the  time  required  for  that  displacement  we 
regard  as  the  average  velocity  for  that  time;  thus,  in  the  last  8  seconds  the  dis- 
placement is  60.6  feet,  and  60.6  -i-  8  =  7.57  ft/sec  is  the  average  velocity  for 
that  time.  (Obviously  a  constant  velocity  of  this  value  would  produce  a  dis- 
placement of  60.6  feet  in  8  seconds.)     In  the  adjoining  table  we  have  listed 

*  For  dimensions  of  a  unit  velocity,  see  Appendix  A. 

118 


Art.  28 


119 


At  (sees.) 

As  (ft.) 

Va  (ft/sec) 

8  to  16  =  8 

60.6 

7-57 

8  to  14  =  6 

42.0 

7.00 

8  to  12  =  4 

26.0 

6.50 

8  to  10  =  2 

12.2 

6.10 

the  displacements  (under  A5)  for  the  intervals  8  to  16,  8  to  14,  etc. 
(under  A^;  and  also  the  average  velocities  (under  Va)  for  these  intervals, 
respectively.  Apparently,  the  average  velocity  for  the  intervals  8  to  9, 
8  to  8|,  8  to  8 J,  etc.,  continues  to  decrease,  approaching  a  definite  limit  as 
the  interval  of  time  approaches  zero.  The  column  of  average  velocities  sug- 
gests that  the  definite  limit  might  be  about  5.8  feet  per  second.  The  exact 
value  of  the  Umit  is  the  rate  at  which  the  ship  was  moving  at  the  time 
8  seconds. 

Summarizing  now:  —  Let  A^  =  distance  traversed  in  any  interval  A/,  and 
Va  =  average  velocity  for  that  interval;  then  in  any  kind  of  rectilinear  motion 


Va  =  ^s/At. 


(2) 


The  value  of  the  velocity  at  a  particular  instant  of  the  interval  is  the  limiting 
value  of  the  average  velocity  for  the  interval,  as  the  interval  is  taken  smaller 
and  smaller  but  always  including  the  particular  instant.  In  the  calculus 
notation,  this  limit  is  ds/dt;  hence  ii  v  =  velocity,  then 

V  =  ds/dt.  (3) 

Here  s  means  the  (varying)  distance  of  the  moving  point  from 
any  fixed  point  in  the  path. 

Formula  (3)  can  be  used  for  finding  the  value  of  v  in  any 
motion  in  which  the  relation  between  s  and  t  is  known.  Thus, 
suppose  that  a  point  is  known  to  move  so  that  its  distance  (in 
feet)  from  the  starting  point  always  equals  four  times  the 
square  of  the  time  (in  minutes)  from  starting,  that  is,  5  =  4  /^; 
then  V  =  ds/dt  —  8t.  This  is  the  general  formula  for  v  in  this 
motion;  that  is,  the  formula  holds  for  all  instants  and  all 
positions  of  the  moving  point.  Thus,  at  the  instant  t  =  2 
minutes,  t)  =  8X2=  16  feet  per  minute;  at  the  position  5  = 
100  feet,  t  —  V  100  -^  4  =  5  minutes  and  z)  =  8  X  5  =  40  feet 
per  minute. 

For  another  example  of  the  use  of  formula  (3)  we  consider  a  "crank  and  con- 
necting-rod mechanism  "  (Fig.  218).  OP  is  the  crank,  mounted  on  its  shaft 
at  O,  PC  is  the  connecting  rod,  C  is  the  crosshead;  (A  is  a  piston  and  AC  the 
piston  rod).     When  the  crank  is  rotated  the  crosshead  is  constrained  to  move 


Fig.  218 


1 20  Chap,  vii 

in  a  straight  line  by  the  guides  G.  We  will  now  find  a  general  formula  for  the 
velocity  of  the  crosshead  (and  piston)  when  the  crank  rotates  uniformly.  Let 
r  =  the  length  of  crank,  /  =  length  of  connecting  rod,  c  =  r/l,  n  =  number 
of  revolutions  of  the  crank  per  unit  time  (assumed  constant),  co  =  angle  in 
radians  described  by  crank  per  unit  time  (co  =  2ivn),  s  =  varying  distance 
of  the  crosshead  from  its  highest  position,  6  =  the  "crank  angle"  QOP,  and 
t  =  time  required  for  the  crank  to  describe  the  angle  6  {=  wt  =  2  irnt). 
Obviously,  there  is  a  definite  relation  between  5  and  d  (or  /),  and  this  relation 
we  need  in  order  to  get  ds/dl  or  v.  When  the  crosshead  is  in  its  highest  posi- 
tion, its  distance  from  0  equals  /  +  r;  hence  for  any  position,  s  =  {I  -\-  r)  — 
CQ  T  OQ,  T  OQ  according  as  the  crank  OP  is  above  or  below  OX.  Now 
CQ  =  (/2  -  r2  sin2  6)'^  =  I  {i  -  c^  sin^  6)^,  and  0Q=  zLr  cos  6;  hence 

5  =  (l -{-  r)  —  I  (i  —  c"^  sin^  d)^  —  r  cos  6. 

Differentiating  the  expression  for  s  with  respect  to  t,  we  get  ds/dt,  or  v;  and 
remembering  that  dd/dt  =  w,  we  can  easily  reduce  the  result  to 

/  c  sin  2  0         \ 

V  2(1  -  c2sin2  0)V 

From  this  general  formula  we  can  get  the  value  of  v  for  any  particular  case. 
Thus,  let  r  =  10  inches,  /  =  30  inches  (then  c  =  \),  and  n  =  100  revolutions 
per  minute  (w  =  2  tt  100  =  628  radians  per  minute).  When  the  crank  is  at 
OPq  say,  d  =  90°  and  the  formula  gives  v  =  6280  inches  per  minute  =  523 
feet  per  minute. 

The  expression  ds/dt  in  equation  (3)  may  be  positive  or  negative;  therefore  v 
must  be  regarded  as  having  the  same  sign  that  ds/dt  has.  Now  ds/dt  is  posi- 
tive when  s  increases  algebraically,  and  ds/dt  is  negative  when  s  decreases 
algebraically;  hence  the  sign  of  the  velocity  of  a  moving  point  at  any  instant 
is  positive  or  negative  according  as  s  is  increasing  or  decreasing  then,  that  is  the 
sign  is  the  same  as  that  of  the  direction  in  which  the  point  is  moving  then. 

When  the  mathematical  relation  between  5  and  t  is  unknown,  then  equation 
(3)  cannot  be  used  to  determine  the  velocity  at  a  particular  instant.     But  if 

the  displacements  of  the  moving  point  are   known  for 
a  number  of  known  intervals  beginning  or  terminating  at 
the  instant  in  question,  then  a  fair  approximation  to  the 
4    S     ^    ^         desired  velocity  can  be  obtained  from  the  values  of  the 
?    "?      I      I         average    velocity   for    those    intervals    as    explained    in 
1 .    !      !      i         the  launching  illustration  preceding.     One  may  determine 
the   limit   of   the    average   velocities    approximately    by 
iG-  219  graphical  methods.     Thus,  in  Fig.  219  we  have  plotted 

the  average  velocities  of  the  launching  example  in  a  manner  which  is  ob- 
vious and  then  joined  the  plotted  points  by  a  smooth  curve;  this  curve 
was  extended,  as  seemed  best,  to  the  vertical  through  point  8.  The  ordinate 
8  A  represents  approximately  the  limit  sought,  that  is  the  velocity  at  the  8th 
second.     Another  graphical  method  is  explained  in  §  2  of  the  following  article. 


Art.  28 


121 


§  2.  Acceleration.  —  A  nonuniform  motion  is  said  to  be  accelerated,  and 
the  moving  point  is  said  to  have  acceleration.  If  the  velocity  changes  uni- 
formly, that  is  by  equal  amounts  in  all  equal  intervals  of  time,  the  motion 
is  uniformly  accelerated;  if  the  velocity  does  not  change  uniformly,  then  the 
motion  is  nonuniformly  accelerated. 

By  acceleration  of  a  moving  point  is  meant  the  rate  at  which  its  velocity  is 
changing.  To  express  the  magnitude  of  any  acceleration  we  must  compare 
that  acceleration  to  some  particular  acceleration  as  a  standard  or  unit.  Any 
rate  of  velocity-change  —  that  of  a  freely  falling  body,  for  example  —  might 
be  taken  as  a  unit  of  acceleration  but  it  is  more  convenient  to  take  the  accel- 
eration of  a  point  whose  velocity  changes  uniformly  by  one  unit  (of  velocity) 
in  one  unit  of  time.  Thus,  we  have  the  foot-per-second  per  second,  the  mile- 
per-hour  per  second,  etc.  And,  abbreviating  the  word  per  as  before,  these 
would  be  written  ft/sec/sec  (also  written  ft/sec  2),  mi/hr/sec,  etc.* 

In  a  unijormly  accelerated  motion  (u.a.m.),  the  acceleration  may  be  computed 
by  dividing  the  velocity-change  which  takes  place  in  any  interval  of  time  by 
the  length  of  the  interval.  Thus,  if  a  =  acceleration,  ^v  =  the  velocity- 
change  and  A^  =  the  interval,  then 

a  =  ^v/^t.  (4) 

In  a  nonuniformly  accelerated  motion  the  rate  of  change  of  the  velocity  is  not 
constant  but  it  varies  continuously  from  instant  to  instant.  To  arrive  at  a 
definite  notion  of  the  value  of  the  rate  or  acceleration  at  a  particular  instant, 
let  us  consider  an  example.  The  adjoining  schedule  gives  values  of  velocity 
and  time  taken  from  a  "starting  test "  of  an  electric  street  railway  car. 


t  =  o 


V  =  o 


I 

2.8 


5-3 


3 

7-7 


4 
9.9 


5  6  7  8  9  10  seconds; 

II. 9        13.7        15.2      16.4      17.3      18.0  miles  per  hour. 


Any  velocity-change  divided  by  the  time  required  for  the  change  we  regard  as 
the  average  acceleration  for  that  time;  thus,  during  the  first  six  seconds  the 
velocity-change  is  13.7  miles  per  hour,  and  13.7  -^  6  =  2.28  miles  per  hour  per 
second  is  the  average  acceleration  for  the  first  six  seconds.  (Obviously  a 
uniform  acceleration  of  this  value  would  produce  in  six  seconds  a  velocity- 
change  of  13.7  miles  per  hour.) 


At  (sees.) 

Av  (mi/hr) 

a„  (mi/hr/sec) 

0  to  6  =  6 

13-7 

2.28 

I  to  6  =  5 

10.9 

2.18 

2  to  6  =  4 

8.4 

2.10 

3  to  6  =  3 

6.0 

2.00 

4  to  6  =  2 

3-8 

1.90 

5  to  6  =  I 

1.8 

1.80 

*  For  dimensions  of  a  unit  acceleration,  see  Appendix  A. 


J 22  Chap,  vn 

In  the  adjoining  table  we  have  listed  the  velocity-changes  (under  Av) 
for  the  intervals  o  to  6,  i  to  6,  2  to  6,  etc.;  and  also  the  average  accel- 
eration (under  da)  for  the  same  intervals.  Obviously  the  average  accelera- 
tion for  the  intervals  5I  to  6,  5!  to  6,  etc.,  continues  to  decrease,  approaching 
a  definite  limit  as  the  interval  approaches  zero.  The  column  of  average 
accelerations  suggests  that  the  definite  limit  might  be  about  1.7  miles  per 
hour  per  second.  The  exact  value  of  the  limit  is  the  rate  of  change  of 
velocity,  or  the  acceleration,  when  t  was  6  seconds.  > 

Summarizing  now:  —  Let  Ay  =  the  velocity-change  in  any  interval  of  time 
At,  and  a«  =  average  acceleration  for  that  interval,  then  in  any  kind  of  recti- 
linear motion 

Qa  =  Av/M.  (5) 

The  true  value  of  the  acceleration  at  a  particular  instant  of  the  interval  is  the 

limiting  value  of  the  average  acceleration  as  the  time  interval  is  taken  smaller 

and  smaller  but  always  including  the  particular  instant;  or  in  the  calculus 

notation 

a  =  dv/dt  =  dh/dt\  (6) 

Formulas  (6),  respectively,  can  be  used  for  finding  the  value  of  a  in  any 
rectilinear  motion  if  the  relation  between  v  and  t  or  5  and  /  are  known.  Thus 
suppose  that  a  point  is  known  to  move  in  a  straight  line  so  that  the  velocity 
(in  miles  per  hour)  always  equals  one-tenth  of  the  square  of  the  time  (in 
seconds)  from  the  start,  that  is  2;  =  o.i  /-;  then  a  =  dv/dt  =  0.2  t.  This  is 
the  general  formula  for  a  in  this  motion ;  for  instance,  at  3  seconds  after  start- 
ing a  =  0.6  miles  per  hour  per  second. 

For  another  example  of  the  use  of  equation  (6),  we  consider  the  motion  of 
the  crosshead  of  the  crank  and  connecting-rod  mechanism  described  in  §  i. 
There  we  found  that 

/  •    /,  1  c  sin  2  ^        \ 

V  2(i-c2sin2  0)V 

Differentiating  this  with  respect  to  /  and  remembering  that  w  is  constant,  we 

get  dv/dt  or 

c  cos  2d  -\-  c^  sin^ d\ 


a  =  rw^  ( cos  6  -f- 


(i-c2sin2  0)^ 


From  this  general  formula,  we  can  get  the  value  of  a  for  any  special  case.  Thus 
as  in  §  I,  let  r  =  10  inches,  /  =  30  inches  (then  c  =  I),  and  n  =  100  revolu- 
tions per  minute  (c<)=27rioo=628  radians  per  minute).  When  the  crank 
is  at  OPo  (Fig.  218),  then  6  =  go  and  the  formula  gives  a  =  —  410  inches 
per  second  per  second.     For  meaning  of  negative  sign,  see  next  paragraph. 

The  expression  dv/dt,  equation  (6),  may  be  positive  or  negative;  therefore  a 
must  be  regarded  as  having  the  same  sign  as  has  dv/dt.  Now  dv/dt  is  positive 
when  the  velocity  increases  algebraically,  and  dv/dt  is  negative  when  the 
velocity  decreases  algebraically;    therefore  the  sign  of  the  acceleration  of  a 


Art.  28 


123 


2      3      4- 

Fig.  220 


6Seci. 


moving  point  at  any  instant  is  positive  or  negative  according  as  the  velocity 
is  increasing  or  decreasing  (algebraically)  then.  Thus  acceleration  is  posi- 
tive or  negative  according  as  positive  or  nega- 
tive velocity  is  being  taken  on  in  the  motion. 
By  direction  of  acceleration  is  meant  the  direc- 
tion of  the  velocity  which  is  being  taken  on. 

When  the  mathematical  relation  between  v  and 
/  (and  5  and  /)  are  unknown,  then  equation  (6) 
cannot  be  used  to  determine  the  acceleration  at 
a  particular  instant.  But  if  the  values  of  the 
velocity  are  known  at  a  number  of  known  in- 
stants near  the  instant  in  question,  then  a  fair  approximation  to  the  accel- 
eration desired  can  be  obtained  from  the  values  of  the  average  acceleration 
for  intervals  beginning  or  terminating  at  the  instant  in  question,  as  explained 
in  the  car-starting  example  preceding.  Fig.  220  shows  a  construction  for 
determining  the  limit  of  the  average  acceleration  in  the  example  referred 
to.  The  ordinate  6  A  represents  the  limit  approximately.  Another  graphic 
method  is  explained  in  the  next  article  under  §  2. 

Note  on  Rate  of  a  Scalar  Quantity.  —  The  foregoing  explanations  of  two  particular  rates 
(velocity  and  acceleration)  will  now  be  generalized  so  that  hereafter  we  will  not  need  to  de- 
rive the  expressions  or  formulas  for  other  rates  which  will  come  up  for  discussion. 

By  a  scalar  quantity  is  meant  one  which  has  magnitude  only,  not  direction  also.  An 
amount  of  money,  the  volume  of  a  thing,  the  population  of  a  city,  etc.,  are  scalar  quantities. 
Let  X  and  y  denote  two  scalar  quantities  which  are  related  to  each  other  so  that  any  change 
in  one  produces  a  change  in  the  other.  If  all  equal  changes  in  x  produce  equal  changes  in  y, 
then  y  is  said  to  vary  uniformly  with  respect  to  x  and  y  is  called  a  uniform  variable.  If  all 
equal  changes  in  x  produce  unequal  changes  in  y,  then  y  is  said  to  vary  nonuniformly  and  y 
is  called  a  nonuniform  variable. 

If  y  is  a  uniform  scalar,  then  the  graph  which  represents  the  relation  between  x  and  y  is  a 
straight  line  obviously,  as  for  example  in  Fig.  221  where  ji  and  y2  respectively  denote  values 
of  y  corresponding  to  xi  and  X2  (values  of  x).  The  meaning  of  "rate  of  y  with  respect  to  a;  " 
or  "a;-rate  of  y  "  is  quite  generally  understood;  it  is  the  change  in  y  per  unit  change  in  x.  The 
value  of  the  rate  is  computed  by  dividing  any  change  in  y  by  the  corresponding  change  in  x. 
Thus,  if  Ax  and  Ay  =  corresponding  changes  in  x  and  y  {x2  -  Xi  and  y2  -  yi),  and  r  =  x-mte 

of  y,  then 

r  =  Ay/Ax. 

Evidently  r  is  the  same  for  all  values  of  x,  that  is,  the  rate  of  a  uniform  scalar  is  constant. 


Fig.  221 

If  y  is  a  nonuniform  scalar  then  the  graph  which  represents  the  relation  between  x  and  y 
is  a  curved  line,  as  for  example  in  Fig.  221  where  vi  and  y.  represent  values  of  y  which  corre- 
spond to  xi  and  x^  respectively.     Any  change  in  y  divided  by  the  corresponding  change  in  x 


124  Chap,  vii 

is  commonly  called  the  average  rate  of  y  with  respect  to  x  for  the  range  xi  —  X\.  Thus,  if 
fa  =  average  x-rate  of  y  for  the  range  Ax  (=  x^  —  x\)  in  x,  then 

Ta  =  ^yl^x. 

The  average  rate  is  represented  by  the  slope  of  the  chord  AB,  for  tan  BAC  =  ^y/^x. 
The  value  of  the  average  x-rate  of  y  depends  on  the  amount  of  the  range  A;r.  It  approaches 
a  definite  value  as  Ax  is  taken  smaller  and  smaller,  x-i  approaching  .vi  for  instance.  This  limit- 
ing value  is  taken  as  the  true  or  instantaneous  rate  of  y  at  the  value  y  =  yi{oix  =  X\).  Thus, 
if  r  =  x-rate  of  y  at  any  value  of  y,  then 

r  =  lim  (Ay /Ax)  =  dy/dx. 

The  x-rate  of  y  at  y  =  yi  is  represented  by  the  limit  of  the  slope  of  the  chord  ^5  as  5  ap- 
proaches A,  that  is,  by  the  slope  of  the  tangent  at  A. 

By  means  of  the  foregoing  formula,  we  can  determine  the  .x-rate  of  y  provided  that  we  know 
the  precise  relation  between  x  and  y,  that  is,  the  equation  y  =  /  (x).  In  case  we  do  not  know 
this  equation  but  do  know  values  of  y  corresponding  to  several  values  of  x,  then  we  can  de- 
termine the  x-rate  of  y  at  one  of  the  values  of  x  approximately.  This  approximate  value  can 
be  obtained  from  the  average  rates  for  ranges  of  x  which  begin  or  terminate  at  the  value  of 
X  for  which  the  rate  is  desired  as  already  explained  in  some  of  the  preceding  examples. 

§  3.  Features  of  a  Motion  Determined  by  Integration.  —  In  the 
preceding  article  we  showed  how  to  determine  the  velocity  from  the  dis- 
tance-time {s-t)  law,  and  the  acceleration  from  the  velocity-time  {v-t)  law. 
The  process,  in  each  case,  is  one  of  differentiation.  By  means  of  the  reverse 
process,  integration,  one  may  determine  the  s-l  from  the  v-t  law,  and  the  v-t 
from  the  a-t  law.     We  explain  further  by  means  of  examples. 

Suppose  that  a  point  moves  in  a  straight  line  according  to  the  law  v  =  dot 
+  4.  In  all  cases  of  rectilinear  motion  v  —  ds/dt,  or  ds  =  v  dt;  hence  in  the 
present  instance,  ds  =  (60  /  +  4)  dl.  Integration  gives  s  =  ^ot'^  -\-  ^t  +  C, 
where  C  is  a  constant  to  be  determined  from  "initial  conditions."  Let  us  sup- 
pose that  s  is  reckoned  from  the  place  where  the  moving  point  is  when  t  =  o, 
or  that  5  =  0  when  /  =  o;  then  substituting  these  (simultaneous)  values  of 
5  and  t  in  the  equation  containing  C,  we  get  o  =  o  -|-  o  +  C,  or  C  =  o.  Hence 
the  s-t  law  is  5  =  30  f^  +  4  /.     We  might  have  integrated  "between  limits," 


thus 


Jds  =    I    (60  /  +  4)  dt,     or    5  =  30  /2  -f  4  /, 
0  t/O 


the  lower  limits  being  the  simultaneous  values  of  s  and  /  from  the  given  initial 
conditions. 

For  another  example,  we  will  suppose  that  a  point  moves  in  a  straight  line 
so  that  a  =  cos  t,  initial  conditions  being  v  =  4  when  /  =  o. 

In  all  cases  of  rectilinear  motion  a  =  dv/dt,  or  dv  =  adt;  hence,  in  this 
instance,  dv  =  cos  tdt.  Integration  gives  t;  =  sin  /  +  C.  Substituting  the 
(initial)  simultaneous  values  of  v  and  /  in  this  equation  we  find  4  =  o  +  C,  or 
C  =  4;  hence  z)  =  sin  /  -f  4  is  the  law  sought.  Or,  integrating  between  limits 
we  get 

COS  tdt,     or    y  —  4  =  sin  t. 


rdv=  r 

t/4  t/0 


Art.  28  125 

If  the  as  law  for  a  motion  is  given,  then  the  v-s  law  can  be  found  by 
integrating  v  dv  =  a  ds,  which  follows  from  a  =  dv/dt  =  {dv/ds)  (ds/dt)  = 
(dv/ds)  V.  Thus,  suppose  that  in  a  rectilinear  motion  a  =  25  +  3,  initial 
conditions  being  ^;  =  10  when  5  =  4;  then 

I  vdv  =    I  (2  s  -{-  7,)ds,     or  ^v^  =  s"^-^  ^s  +  C. 

Initial  values  substituted  in  the  last  equation  give  C  =  22,  and  hence  ^  v^  =  s^ 
+  35+  22.     And  in  any  rectilinear  motion 


ads  =   I     vdv  =  ^  {v^  —  v^), 


where  Vi  and  V2  are  values  of  the  velocity  when  s  =  Si  and  5  =  ^2  respectively. 
The  formulas  a  =  dv/dt  and  v  =  ds/dt  can  be  used  to  get  "time."     These 
can  be  written  dt  =  {i/a)  dv  and  dt  =  (i/v)  ds;  hence  by  integration 

-  dv,     and     4  —  ^1  =   /      -ds. 
V,   a  Js^   V 

These  respectively  give  the  time  required  for  v  to  change  from  vi  to  V2,  and  for 
s  to  change  from  Si  to  ^2. 

Uniformly  Accelerated  Motion.  —  Let  a  =  the  value  of  the  (constant)  accel- 
eration, and  Vo  =  the  velocity  at  the  instant  from  which  time  is  reckoned,  and 
So  =  the  distance  of  the  moving  point  from  the  origin  at  that  instant.  (Some- 
times Vo  and  So  are  called  initial  velocity  and  initial  distance,  respectively.) 
Since  a  is  constant,  integration  of  a  =  dv/dt  gives  at  once  v  =  at-\-  Ci,  and 
from  the  initial  conditions  (v  =  Vo  when  /  =  o),  Ci  =  Vo,  hence 

V  =  at  -{-  Vq.  (i) 

From  V  =  ds/dt  =  at  -{-  Vo  we  find  by  integration  that  s  =  I  at"^  -\-  Vot  +  C2, 
and  the  initial  conditions  {s  =  So  when  t  =  o)  make  C2  =  So',  hence 

5  =  I  a/2  +  Vot  +  So.  (2) 

Eliminating  t  between  (i)  and  (2)  we  find  that 

2a(s  —  So)  =  v^  —  vo"^.  (3) 

If  the  initial  velocity  and  distance  =  o,  then 

v  =  at,     5  =  1  at^,     and     2  as  =  v^.  (4) 

Although  uniformly  accelerated  motions  are  important  practically,  the 
student  is  advised  not  to  make  a  special  effort  to  memorize  the  foregoing 
formulas  (i,  2,  3,  and  their  special  forms,  4).  But,  if  he  will  memorize  them, 
then  he  should  also  remember  that  they  are  for  a  special  motion,  constant 
acceleration.  All  students  ought  to  be  able  to  discuss  a  uniformly  accelerated 
motion  nonmathematically  —  by  means  of  elementary  notions  somewhat  as 
in  the  following  example:  The  velocity  of  a  certain  train  can  be  reduced  by 
braking  from  40  to  20  miles  per  hour  in  a  distance  of  1600  feet.     In  what  dis- 


126 


Chap,  vii 


tance  would  braking  stop  the  train  from  40  miles  per  hour,  supposing  the 
retardation  to  be  the  same  at  all  velocities?  Since  the  velocity  changes  uni- 
formly, the  average  velocity  during  the  reduction  from  40  to  20  miles  per  hour 
equals  one-half  of  4c  +  20  or  30  miles  per  hour;  and  the  time  required  for  the 
reduction  of  velocity  or  travel  of  1600  feet  (=  0.303  miles)  is  0.303  -^  30 
=  0.0101  hours,  or  36.4  seconds.  The  time  required  to  stop  the  train  from  40 
miles  per  hour  would  be  twice  36.4  or  72.8  seconds;  and,  inasmuch  as  the 
average  velocity  during  the  stoppage  would  be  one-half  of  (40  -f  o)  =  20 
miles  per  hour  or  29.3  feet  per  second,  the  distance  travelled  in  the  72.8  seconds 
would  be  29.3  X  72.8  =  2133  feet. 

29.   Motion  Graphs 

The  features  of  rectilinear  motion,  discussed  in  the  preceding  article,  can  be 
represented  nicely  by  certain  curves  described  in  the  following: 

A  distance-time  (s-t)  graph  for  any  motion  is  a  curve  drawn  "upon"  a 
pair  of  rectangular  reference  axes  so  that  the  coordinates  of  any  point  on  the 

curve  represent  corresponding,  or  simulta- 
neous, values  of  s  and  t,  where  t  =  the  time 
elapsed  from  some  instant  of  reckoning  (usu- 
ally taken  at  the  instant  of  starting),  and  5 
=  the  distance  of  the  moving  point  from 
some  fixed  point  chosen  as  origin  (usually 
taken  at  the  place  of  starting).  Fig.  222  is 
the  s-t  graph  for  the  launching  mentioned 
in  §  I  of  the  preceding  article.  Since  the  slope  of  the  s-t  graph  is  proportional 
to  ds/dt  and  v  =  ds/dt,  the  slope  at  any  point  of  the  graph  represents  the 
velocity  at  the  corresponding  instant,  according  to  some  scale.  The  slope 
scale  depends  on  the  scales  used  for  plotting  the  s-t  graph.  Thus,  in  Fig.  222 
the  scales  are  i  inch  of  ordinate  =  100  feet  and  i  inch  of  abscissa  =  10  seconds, 
hence,  a  slope  of  unity  =  100  (feet)  ^10  (seconds)  =  10  (feet  per  second). 
Thus,  the  velocity  at  /  =  8  seconds,  where  the  slope  h  BC  -^  AC  =  0.54,  is 
5.4  feet  per  second.  Instead  of  interpreting  the  slope  in  this  way,  that  is  by 
a  slope  scale,  we  might  determine  the  velocity  as  follows:  draw  the  tangent 
line  at  the  point  A  in  question,  drop  a  per- 
pendicular from  any  point  B  in  the  tangent 
to  the  horizontal  through  A,  measure  CA 
and  CB  according  to  the  proper  scales  and 
compute  the  ratio  BC  h-  AC  (as  measured); 
this  ratio  equals  the  desired  velocity.  Thus, 
in  Fig.  222,  AC  =  5  seconds,  CB  =  27  feet, 
and  ^  =  27  ^  5  =  5.4  feet  per  second.* 

*  Several  instruments  have  been  devised  recently  for  drawing  a  tangent  to  a  plane  curve. 
A  very  simple  one  is  represented  in  Fig.  223.  It  consists  of  a  metal  straight-edge  A  with  a  por- 
tion of  one  side  polished  to  a  mirror.     OB  represents  a  curve  on  a  piece  of  paper  across  which 


16  Sees. 


Fig.  222 


Fig.  223 


Art.  29 


127 


The  velocity-time  (v-t)  graph  for  any  rectilinear  motion  is  a  curve  drawn 
upon  a  pair  of  rectangular  reference  axes  so  that  the  coordinates  of  any  point 
of  the  curve  represent  corresponding,  or  simultaneous,  values  of  the  velocity 
V  and  time  /.     The  curve  in  Fig.  224  is  a  v-t  graph  for  the  car-starting  test 


Fig.  224 


I05ec&, 


mentioned  in  §  2  of  the  preceding  article.  The  slope  of  the  v-t  graph  at  any 
point  represents  the  acceleration  at  the  corresponding  instant.  To  actually 
determine  the  acceleration  from  the  graph,  the  slope  must  be  interpreted  by 
proper  scale  or  be  computed  in  a  manner  analogous  to  that  explained  in  the 
foregoing  under  distance-time  graph.  Thus,  at  the  fifth  second,  the  accelera- 
tion is  represented  by  the  slope  of  the  tangent  at  A;  since  AC  =  2.5  seconds 
and  CB  =  4.8  miles  per  hour,  the  acceleration  =  4.8  -^  2.5  =  1.92  miles  per 
hour  per  second. 

The  "area  under  the  curve  "  (between  the  curve,  the  time  axis,  and  any  two 
ordinates)  represents  the  displacement  for  the  interval  of  time  represented  by 
the  distance  between  the  ordinates.  Proof:  Let  m  =  velocity  scale-number 
and  n  =  time  scale-number,  that  is 

unit  ordinate  (inch)  =  m  units  of  velocity  (feet  per  second) ; 
unit  abscissa  (inch)  =  n  units  of  time  (seconds). 

Thus,  let  X  and  y  be  the  lengths  (inches)  of  the  coordinates  of  any  point  P  of 
a  v-t  curve  (Fig.  225) ;  then  the  corresponding  values  of  v  and  /  are  my  and  nx. 

the  straight-edge  is  laid  at  random  but  so  that  a  portion  of  the  curve  is  reflected  from  the 
mirror.  The  image  CD  and  the  curve  CO  are  not  smoothly  continuous;  there  is  a  cusp  at  C. 
But  if  the  instrument  be  turned  about  C  until  the  cusp  disappears,  the  curve  merging  smoothly 
into  its  image,  then  the  straight-edge  A  is  normal  to  the  curve  OB  at  C.  Having  located  the 
normal  at  C,  it  is  easy  to  draw  the  tangent.  The  principle  of  this  instrument  is  the  basis  of 
Wagener's  derivator  (see  Gramberg's  Tcchnische  Messungcn)  by  means  of  which  the  slope  of  a 
curve  at  any  point  can  be  read  directly,  without  drawing  the  tangent  or  normal.  An  auto- 
graphic form  of  (mirror)  derivator  has  been  devised  by  A.  Elmendorf  (see  Sci.  Am.  Suppl. 
for  Feb.  12,  1916). 

Guillery's  "aphegraphe"  is  another  instrument  for  drawing  a  tanf^ent  to  a  plane  curve. 
A  metal  strip  or  batten  must  first  be  fitted  to  the  cuive  before  the  instrument  proper  can  be 
applied.  For  full  description  of  the  aphegraphe,  see  Mem.  Soc.  Ing.  Civ.  de  France,  Bull,  for 
.■\pril,  1911,  where  M.  Guillery  also  explains  how  he  applied  his  instrument  to  determine  the 
acceleration-time  curves  for  several  mechanisms,  and  in  particular  the  a-t  curve  for  the  "tup" 
of  an  impact  testing  machine  during  a  blow. 


128 


Chap,  vn 


Further  let  h  and  k  =  the  times  corresponding  to  Xi  and  3C2  and  to  Si  and  52,  the 
values  of  s  (space) ;  and  A  =  area.    Then 

/»^j     ,          rkv  dt        I     r'2     , 
=    1     ydx  =    I =  —    /     vdt  = 


A 


S2  —  S1 


mn 


(see  preceding  article) ;  and  A  {mn)  =  s^  —  Si.  Hence 
{mn)  is  the  scale-number  for  interpretating  the  area. 
Thus,  in  Fig.  224,  one-inch  ordinate  =  20  miles  per 
hour  =  29.3  feet  per  second,  and  one-inch  abscissa  = 
Fig.  225  ^   seconds;    hence  one  square  inch  =  29.3    (feet   per 

second)  X  5  (seconds)  =  146.5  feet.  The  area  may  be  interpreted  more 
directly  by  multiplying  the  average  ordinate  measured  by  the  scale  of 
ordinates  (hence  equal  to  the  average  velocity  for  the  time  interval)  by  the 
length  of  the  interval.  Thus,  in  Fig.  224  the  average  ordinate  represents  10.9 
miles  per  hour  =16  feet  per  second,  and  the  time  interval  is  10  seconds, 
hence  the  displacement  is  160  feet. 

The  acceleration-time  {a-t)  graph  for  any  rectilinear  motion  is  a  curve 
drawn  upon  a  pair  of  rectangular  reference  axes  so  that  the  coordinates  of  any 
point  of  the  curve  represent  corresponding,  or  simultaneous,  values  of  the  ac- 
celeration a  and  the  time  /.  The  "area  under  the  curve"  represents  the 
velocity-change  for  the  time  interval  represented  by  the  distance  between  the 
ordinates.     For  the  area  under  the  curve  is  given  by 

a  dt,     and    Vi  —  V\=    I    adt 
h  Jh 

(see  preceding  article).  To  determine  the  numerical  value  of  the  velocity- 
change,  the  area  must  be  interpreted  by  scale  or  be  computed  in  a  manner 
analogous  to  that  explained  in  the  foregoing  under  velocity-time  graph. 

The  velocity-distance  {v-s)  graph  for  a  rectilinear  motion  is  a  curve  drawn 
upon  a  pair  of  rectangular  axes  so  that  the  coordinates  of  any  point  of  the 
curve  represent  corresponding,  or  simultaneous, 
values  of  the  velocity  v  and  distance  s.  Fig.  226 
is  the  v-s  graph  for  an  air-brake  test  on  a  pas- 
senger train.*  The  subnormal  at  any  point  of  the 
graph  represents  the  acceleration  at  the  corre- 
sponding instant.  For,  any  subnormal  as  BC  is 
given  by  ^C  tan  BAC  =  vdv/ds,  and  from  the 
preceding  article  a  =  dv/dt  =  {dv/ds)  {ds/dt)  = 
vdv/ds;  hence  BC  =  a.  To  actually  determine 
the  value  of  a  from  a  subnormal  we  must  use  the 
proper  scale,  depending  on  the  scales  used  for  plotting  the  v-s  graph.  For  Fig. 
226  one-inch  ordinate  =  50  miles  per  hour,  and  one-inch  abscissa  =  1000  feet 
=  0.19  mile;  hence  the  subnormal  scale  is  one  inch  =  50'  -r-  0.19  =  13,150  miles 
per  hour  per  hour  =  3.65  miles  per  hour  per  second.  The  subnormal  BC 
*  "Air-brake  Tests  —  Westinghouse."     Page  297. 


200    400    60O    800     1000 

Fig.  226 


Art.  29 


129 


=  0.72  inch;  hence  the  (negative)  acceleration  at  A  (when  the  train  had  made 
600  feet  from  the  place  where  braking  began)  was  0.72  X  3.65  =  2.63  miles 
per  hour  per  second. 

The  acceleration-distance  (as)  graph  for  a  rectilinear  motion  is  a  curve 
drawn  upon  a  pair  of  rectangular  axes  so  that  the  coordinates  of  any  point  on 
the  graph  represent  corresponding,  or  simultaneous,  values  of  a  and  s.  "Area 
under  the  curve  "  (between  the  curve,  the  5  axis,  and  ordinates  ai  and  a2) 
represents  one-half  the  change  in  the  velocity-square  corresponding  to  the 
change  o^  —  ai  or  ^2  —  ^i.     For,  the  area  is  given  by 


ads  =   I 


vdv  =  ^  {v^  —  v^). 


motion  is 


The.  reciprocal  acceleration-velocity  {- -v\  graph  for  a  rectilinear 

a  curve  drawn  upon  rectangular  axes  so  that  the  coordinates  of  any  point 
on  the  curve  represent  corresponding,  or  simultaneous,  values  of  i/a  and  v. 
"Area  under  the  curve"  (between  the  curve,  the  v  axis,  and  ordinates  i/ai 
and  1/C2)  represents  the  time  required  for  the  acceleration  to  change  from 
ai  to  02,  or  velocity  from  Vi  to  V2-     For,  the  area  is  given  by 

^dv=   I     dt=  k-  k. 
I),    d  Jh 


The  reciprocal  velocity-distance  {--s\  graph   for  a  rectilinear   motion 


IS   a 


curve  drawn  upon  rectangular  axes  so  that  the  coordinates  of  any  point  on 
the  curve  represent  corresponding,  or  simul- 
taneous, values  of  i/v  and  5.  "Area  under 
the  curve"  (between  the  curve,  the  s  axis, 
and  ordinates  i/i'i  and  i/v^)  represents 
the  time  required  for  the  velocity  to  change 
from  V\  to  V2.     For,  the  area  is  given  by 


I  sec. 


dt  =  t^  —  tu 


I  sec. 


Example.  —  A  mechanism  is  to  be  de- 
signed for  producing  a  rectiUnear  motion 
whose  acceleration-time  graph  is  shown 
in  Fig.  227.  There  are  three  distinct 
laws  of  acceleration.  In  the  first  and 
last  quarter  seconds  the  acceleration  is 
constant  and  equals  16  feet  per  second 
per  second;  in  the  second  quarter  the  ac- 
celeration decreases  uniformly  from  16  to 
—  48;  and  in  the  third  it  increases  uniformly  from  —48  to  16.  Preliminary 
to  the  design  it  is  necessary  to  find  the  distance-time  law;  this  we   proceed 


I'A       ^/8        "z        %       % 

Figs.  227,  228,  229 


F 

I  sec. 


i^o  Chap,  vii 

to  do,  but  first  we  get  the  velocity-time  and  distance-time  graphs  approxi- 
mately. 

During  the  first  quarter  of  a  second  the  velocity  changes  uniformly,  and 
the  change  is  i6  X  j  =  4  feet  per  second;  and  if  the  initial  velocity  is  zero, 
then  OA  (Fig.  228)  is  the  velocity- time  graph  for  the  first  quarter  second. 
Since  the  velocity  changes  uniformly  in  the  first  quarter  second,  the  average 
velocity  equals  |  (o  -f  4)  =  2  feet  per  second,  and  the  displacement  during 
the  quarter  =2X5  =  ^  foot.  If  the  initial  distance  is  zero  then  0  and  A 
(Fig.  229)  are  points  on  the  distance-time  graph.  In  a  similar  way  interme- 
diate points  could  be  computed. 

In  the  second  quarter  the  acceleration  varies  uniformly.  The  average 
acceleration  for  the  interval  from  |  to  t^  second  is  8  feet  per  second  per  second; 
hence  the  velocity-change  for  that  interval  is  8  X  yV  =  2  foot  per  second,  and 
the  velocity  at  /  =  /^  is  4  -|-  ^  =  4.5  feet  per  second,  and  B  (Fig.  228)  is  a  point 
on  the  velocity-time  graph.  In  a  similar  way,  C,  D,  and  intermediate  points 
could  be  determined.  The  portion  AD  is  curved,  and  the  average  velocity 
for  any  interval  cannot  be  ascertained  so  simply.  But  estimating  the  average 
ordinate  for  the  third  eighth  of  a  second  to  be  4.4,  then  the  displacement  for 
that  interval  is  4.4  X  |  =  0.55  feet,  and  C  (Fig.  229)  is  another  point  on  the 
distance- time  graph.  In  a  similar  way  we  might  determine  other  points 
approximately.  Determination  of  the  graphs  for  the  third  and  fourth  quarter 
seconds  by  this  method  presents  no  difficulties,  so  we  pass  on  to  a  second 
(mathematical)  determination  of  the  graphs. 

In  the  first  quarter,  dv/dt  =  16,  or  dv  =  16  dt;  hence  v  -^  i6t+  C.  But 
in  accordance  with  initial  conditions  assumed,  v  =  o  when  /  =  o;  hence 
C  =  o,  and  z;  =  16  /  is  the  equation  of  the  velocity-time  graph  for  the  first 
quarter.  From  that  equation  we  find  for  t  =  I,  v  =  4  as  before.  Since 
V  =  ds/dt,  ds  =  vdt  =  16  tdt,  and  s  =  Sf  +  C.  In  accordance  with  initial 
conditions  assumed,  s  =  o  when  /  =  o;  hence  C  =  o,  and  5  =  8  ^  is  the  equa- 
tion of  the  distance-time  graph  for  the  first  quarter.  From  that  equation  we 
find  for  /  =  I,  5  =  I  as  before;  at  /  =  i  5  =  |  foot;  etc. 

In  the  second  quarter,  c  =  80  -  256  /,  equation  of  AD  (Fig.  227);  hence 
dv  =  (80  -  256  /)  (//  or  t'  =  80  /  -  128  /2  -f  C.  We  found  that  v  =  4  when 
t  =  i;therefore4  =  80  X  i  -  128  X  tV  +  C,  orC  =  -  8,  andz;  =&ot- 128^-8 
is  the  equation  of  the  velocity-time  graph  for  the  second  quarter.  Continu- 
ing, 


ds/dt  =  80  /  -  128  /2  -  8,    or   5  =  40  /^  -  42I  /^  -  S  /  -h  C;   but 


S  —   2 

when  /  =  I,  hence  C  =  f ,  and  5  =  40  /^  -  42!  /^  -  8  /  +  f  is  the  equation 
of  the  distance-time  graph  for  the  second  quarter. 

The  equations  of  the  graphs  for  the  remaining  quarters  could  be  obtained 
in  a  similar  way.  Care  must  be  taken  in  determining  the  constants  of  inte- 
gration; use  no  value  of  /  (and  corresponding  value  of  v  or  s)  which  does  not 
fall  within  the  period  to  which  the  equation  under  consideration  pertains. 
The  remaining  equations  are  — 


rtRT.  30 


131 


4ft.persec.persec. 
1  16  sees. 


24 


For  the  third  quarter  For  the  fourth  quarter 

a  =  —  176  +  256 1  c  =  16, 

2,  =  —  176  /  +  128  /2  +  56  z;  =  16  /  —  16, 

5=  -88/2  +  421^+56/- 10  5=8/2-16/+  8. 

Graphs  for  Uniformly  Accelerated  Motion.  —  Fig.  230  shows  the  acceleration- 
time  graph  for  a  rectilinear  motion;    in  the  first  six  seconds  a  =  4  feet  per 

second  per  second,  in  the  next  ten  seconds  a  ||  o 
and  in  the  last  8  seconds  a  =  —  3  feet  per  second 
per  second  (the  negative  sign  meaning  retardation). 
Fig.  231  shows  the  corresponding  velocity-time 
graph,  it  being  assumed  that  there  is  no  initial 
velocity.  Fig.  232  shows 
the  corresponding  dis- 
tance-time graph,  initial 
distance  being  taken  as 
zero.  Fig.  233  shows 
the  as  and  v-s  graphs; 
AB-CD-EF  is  the  former, 
and  OGHJ  is  the  latter. 


6  sees. 


O 


■3 


6  sees. 


I6secs. 


24 


15sec5. 


5 


\(  72  M- 


Figs.  230,  231,  232 


|D 

IE IF 

I 1 

■  —  240' ->k-96'->l 

Fig.  233 


30.   Simple  Harmonic  Motion  and  a  Similar  One 

§  I.  Simple  Harmonic  Motion  (S.H.M.).  —  If  a  point  moves  uniformly 
along  the  circumference  of  a  circle  then  the  motion  of  the  projection  of  that 
point  on  any  diameter  is  called  a  simple  harmonic  motion.  Obviously  the 
projection  {Q)  moves  to  and  fro  in  its  path,  and  travels  the  length  of  the 
diameter  twice  while  the  point  (P)  in  the  circumference,  goes  once  around. 
By  amplitude  of  the  s.h.m.  is  meant  one-half  the  length  of  the  path  of  Q,  equal 
to  the  radius  of  the  circle,  '^y  frequency  oi  the  s.h.m.  is  meant  the  number  of 
complete  (to  and  fro)  oscillations  of  the  moving  point  Q  per  unit  time,  equal  to 
the  number  of  excursions  of  P  around  the  circumference  per  unit  time.  By 
period  of  the  s.h.m.  is  meant  the  time  required  for  one  complete  to  and  fro 
oscillation  of  the  moving  point  Q,  equal  to  the  time  required  for  one  excursion 
of  P  around  the  circle.  By  displacement  of  the  moving  point  Q  is  meant  its 
distance  from  the  center  of  the  path;  it  is  regarded  as  positive  or  negative 
according  as  Q  is  on  the  positive  or  negative  side  of  the  center. 

Let  us  now  consider  a  simple  harmonic  motion  to  ascertain  approximately 
its  nature.  Suppose  that  the  circle  (Fig.  235)  to  be  the  path  of  P,  and  the 
vertical  diameter,  say,  the  path  of  Q.  The  y-t  (space-time)  and  the  y-d 
graphs  for  the  motion  of  Q  can  be  constructed  very  easily.  We  mark  any 
number,  say  sixteen,  equidistant  positions  of  P,  and  number  them  consecu- 
tively and  also  the  positions  of  Q  to  correspond  (Fig.  234).  Then  on  an  exten- 
sion of  the  horizontal  diameter  we  lay  off  any  convenient  length  oT  to  represent 
360°  or  the  period,  and  divide  oT  into  sixteen  equal  parts  numbering  the  points 
of  division  as  shown.     Finally  we  project  points  o,  i,  2,  etc.,  of  the  circle  upon 


132 


Chap,  vu 


the  verticals  through  the  corresponding  points  o,  i,  2,  etc.,  of  oT.  These 
projections  are  on  the  y-t  or  y-d  graph.  The  slope  of  the  graph  at  any 
point  represents  the  velocity  of  Q  at  the  corresponding  instant;   hence  the 


_2/' 

"^  i  ^ 

\6 

A  : 

i   '.\w 

0         2 

4 

6        8\    ;     i;0    ;     1 

2 

uo"  e 


'\<0 


12 


Fig.  234 


velocity  of  <3  is  greatest  at  the  middle  of  its  path,  and  equals  zero  at  the  ends 
of  the  path.  The  arcs  o-i,  1-2,  2-3,  etc.,  are  equal,  and  are  therefore 
described  by  the  moving  point  P  in  equal  intervals  of  time.  The  lengths 
o-i,  1-2,  2-3,  etc.,  in  the  diameter  are  described  by  Q  in  the  same  equal 
intervals  of  time;  hence  the  average  velocities  of  Q  for  those  intervals  are  pro- 
portional to  those  distances.  For  comparison,  the  distances  were  laid  off 
from  O;  OA  =  o-i,  OB  =  1-2,  OC  =  2-3,  and  OD  =  3-4.  Carefully  com- 
paring these  distances,  we  see  t^at  the  numerical  difference  between  succes- 
sive average  velocities  increases;  hence  the  acceleration  increases  in  value 
as  Q  moves  from  o  to  4.  In  fact  the  acceleration  is  zero  when  Q  is  at  the 
middle,  and  greatest  when  at  either  end  of  its  path  (proved  below). 
We  now  examine  s.h.m.  more  carefully,  using  the  following  notation:  — 

r  =  amplitude  (radius  of  the  circle), 

n  =  frequency, 

o)  =  2  TTW  (abbreviation),  co  being  angle  in  radians  swept  out  per  unit  time 

by  OP, 
X,  y,  or  s  =  displacement, 

t  =  time  after  some  convenient  origin  as  described  later, 
V  =  velocity  of  the  s.h.m.  at  the  time  /, 
a  =  acceleration  of  the  s.h.m.  at  the  time  t. 

When  time  t  is  reckoned  from  the  instant  when  Q  was  at  middle  of  its  path  and 
moving  in  positive  direction.  —  Suppose  the  circle  (Fig.  235)  to  be  the  path  of  P, 

which  moves  in  sense  indicated  by  arrow,  and  let  us 
consider  the  motion  of  the  projection  of  P  on  the  verti- 
cal diameter,  from  now  on  called  V  instead  of  Q.  Let 
6  =  angle  XOP;  then,  since  t  is  time  elapsed  since  P 
was  a.t  X,  6  =  2  irnt  =  co/  and  dd/dt  =  co.  It  is  plain 
from  the  figure  that  y  =  r  sin  0;  and  since  v  =  dy/dt  =  r 
(dd/dt)  cos  d, 

Fig.  235  V  =  rw  cos  9  =  ru  cos  co/.  (i) 

These  are  (general)  formulas  for  v  in  terms  of  6  and  /  respectively. 


Art.  30 


133 


Since  cos  9  =  sin  (d  -\-  ^ir),  v  =  m  sin  (5  +  §  tt).  This  formula  for  v  sug- 
gests an  easy  method  for  drawing  a  v-d  graph,  showing  how  the  velocity 
varies  with  d,  and  hence  with  /.  First  we  draw  an  auxiliary  circle  with  radius 
equal  to  2  Trn  according  to  any  convenient  scale;  divide  the  circumference  into 
any  convenient  number  of  equal  parts,  as  sixteen;  and  number  the  points  of 
division  as  in  Fig.  236,  that  is  90°  ahead  of  the  numbers  in  Fig.  234.     On  an 


Fig.  236 


extension  of  the  horizontal  diameter  we  lay  off  oT  to  represent  360°,  and  sub- 
divide this  into  the  same  number  of  equal  parts  (sixteen),  numbering  as  shown; 
then  01,  02,  etc.,  represent  6  =  22^°,  6  =  45°,  etc.  Finally  we  project  points 
o,  I,  2,  etc.,  of  the  circle  toward  the  right  to  meet  corresponding  vertical  lines 
through  points  o,  i,  2,  etc.,  of  oT.  These  points  of  meeting  are  on  the  v-9 
graph,  for  the  coordinates  of  any  point  on  the  curve  are  corresponding,  or 
simultaneous,  values  of  d  and  rw  sin  (^  +  |  tt)  ,  or  v. 

Inspection  of  the  v-d  graph  verifies  what  was  said  about  the  acceleration. 
It  shows  clearly  that  the  velocity  of  the  moving  point  V  (Fig.  235)  changes  more 
rapidly  when  V  is  near  the  ends  of  its  path  than  when  near  the  center;  hence 
the  acceleration  of  V  is  greater  near  the  ends  than  near  the  center.  Since  the 
v-d  graph  is  also  a  v-t  graph,  the  slopes  of  the  graph  represent,  to  proper 
scale,  values  of  the  (varying)  acceleration.  The  curve  is  steepest  when  6  =  90° 
and  270°  (when  V  is  at  the  ends  of  its  path),  and  horizontal  when  0  =  o  and 
180°  (when  V  is  at  the  center  of  its  path);  hence  again  the  acceleration  is 
greatest  at  the  ends  of  the  path,  and  zero  at  the  center.  When  the  moving 
point  V  is  approaching  the  center  of  its  path  —  from  either  side  —  then  V  is 
getting  up  speed,  and  hence  the  acceleration  of  V  is  directed  toward  the  center; 
when  V  is  receding  from  the  center,  then  V  is  slowing  down,  and  hence  the 
acceleration  is  directed  toward  the  center.  Therefore  the  acceleration  is 
always  directed  toward  the  center. 

A  general  formula  for  acceleration  in  a  s.h.m.  will  now  be  derived.  We  take 
the  motion  of  V  (Fig.  235)  for  that  purpose,  and  let  a  —  the  acceleration  at  any 
time  /.     Now  0  =  dv/dt,  v  =  rco  cos0,  and  dd/dl  —  co;  hence 

a=  —  rorsind  =  —  ror  sin  cct.  (2) 

These  are  (general)  formulas  for  a  in  terms  of  d  and  t  respectively. 

Since  sin  0  =  —  sin  {6  +  tt),  a  =  rw"  sin  (d  -\-  ir).  This  last  formula  sug- 
gests an  easy  method  for  drawing  the  a-9  graph,  showing  how  a  varies  with 


134 


Chap,  vii 


6,  and  hence  also  with  t.  First  we  draw  an  auxiUary  circle  (Fig.  237)  with 
radius  equal  to  rw-  according  to  any  convenient  scale;  divide  the  circumfer- 
ence into  any  convenient  number  of  equal  parts,  say  sixteen;    and  number 


Fig.  237 

them  as  in  the  figure,  that  is  180°  ahead  of  the  numbers  in  Fig.  234.  On  an 
extension  of  the  horizontal  diameter  we  lay  off  OT  to  represent  360°,  and  sub- 
divide OT  into  sixteen  equal  parts  numbering  as  shown;  then  01,  02,  etc., 
represent  d  =  22^°,  6  =  45°,  etc.  Finally,  we  project  points  o,  i,  2,  etc.,  of 
the  circle  horizontally  to  meet  the  corresponding  vertical  lines  through  points 
o,  I,  2,  etc.,  of  the  Hne  OT.  These  points  of  meeting  are  on  the  a-d  graph, 
for  the  coordinates  of  any  point  on  the  curve  are  corresponding,  or  simulta- 
neous, values  of  9  and  rw^  sin  (6  -\-ir),  or  a. 

In  Fig.  238  the  foregoing  described  distance,  velocity,  and  acceleration  graphs 
are  superimposed;  the  solid  curve  is  the  y-d  graph,  the  dashed  curve  the 
v-d  graph,  and  dot-dash  curve  is  the  a-d  graph. 


Fig.  238  Fig.  239 

Time  dated  from  the  instant  when  Q  was  at  the  positive  end  of  its  path.  —  We 
might  continue  to  regard  the  s.h.m.  as  taking  place  in  the  vertical  diameter 
of  Fig.  235,  reckoning  time  from  the  instant  when  P  was  at  Y.  It  will  be  more 
convenient  to  consider  the  motion  of  the  projection  of  P  on  the  horizontal 
diameter;  then  we  measure  6  and  /  as  before.     It  is  easy  to  show  that 

x=rcosd  =  rcoso:t;  v=  —r(j}sm6=  — rcosinw/;  a=  —  rw^  cos  0  =  —  rco^  cos  w/. 
Time  dated  from  the  instant  when  Q  was  at  some  intermediate  point.  —  Let  / 
be  reckoned  from  the  instant  when  P  (Fig.  239)  was  at  some  point  as  Pq,  and 
let  6  =  PoOP  and  e  =  XOPo-  This  latter  angle  is  called  angle  of  lead;  but 
angle  of  lag  when  Po  is  below  OX.  Now,  XOP  =  0  +  e  =  co/  +  e.  In  the 
s.h.m.  executed  by  V, 

J  =  r  sin  (^  +  e) ;     v  =  rco  cos  (0  -f-  e);     a  =  —  r(j?  sin  (0  +  e). 
In  the  s.h.m.  executed  by  ZT, 

x  =  r  cos  (0  -F  e) ;     z;  =  —  rw  sin  (5  +  e) ;     a=  —  to?  cos  (0  +  e). 


Art.  30 


135 


Formulas  jor  Velocity  and  A  cceleration  in  Terms  oj  Displacement.  —  These 
do  not  depend  on  the  way  in  which  time  is  reckoned.  Referring  to  the  fore- 
going formulas  we  see  that 

V  =  0}  Vr^  —  s"^  =  ior  Vi  —  {s/rY, 


a 


co'^s. 


where  5  stands  for  displacement  x  or  y. 

The  graph  oi  v  =  0:  Vr^  —  s~  is  the  velocity-displacement  graph  for  any 
s.h.m.;  it  is  an  ellipse.  Fig.  241  shows  that  graph  for  the  motion  of  the  pro- 
jection of  P  on  the  horizontal  diameter  of  the  circle.  When  P  is  where  indi- 
cated say,  the  velocity  of  H  is  represented  by  the  ordinate  HV.  The  graph  of 
a  =  —  co-5  is  the  acceleration-displacement  graph ;  it  is  a  straight  line.  The 
diagonal  line  in  the  figure  is  the  a-s  graph  for  the  motion  of  H.  The  acceler- 
ation of  H  is  represented  by  the  ordinate  HA. 


if        Hi    H 

oIh  :h      jx 

\        ; 

fc; 

— ^ 

%i. 

Fig.  241 


Fig.  242 


Mechanism  for  Producing  a  Simple  Harmonic  Motion.  —  The  mechanism 
represented  in  Fig.  242  consists  of  a  crank  C  and  a  slotted  slider  S.  The  slider 
is  constrained  by  fixed  guides  G  so  that  it  can  be  moved  to  and  fro  only  (verti- 
cally in  this  figure).  The  crank-pin  P  projects  through  the  slot  of  the  slider; 
hence  if  the  crank  be  turned,  the  crank-pin  presses  against  and  moves  the 
slider.  If  the  crank  be  turned  uniformly  then  every  point  of  the  slider  exe- 
cutes a  simple  harmonic  motion. 

§  2.  Crank  and  Connecting-rod  Mechanism  (Fig.  243). — When  the 
crank  is  rotating  uniformly,  the  motion  of  the  crosshead  (and  piston)  resem- 


FiG.  243 


bles  a  simple  harmonic  one  quite  closely,  as  will  be  shown  presently.  Exact 
formulas  for  the  position,  velocity,  and  acceleration  of  the  crosshead  for  any 
position  of  the  crank  were  derived  in  Art.  28.     The  formulas  are  not  simple. 


136 


Chap,  vii 


The  following  approximate  formulas  (i,  2,  and  3)  are  simpler  and  quite  accu- 
rate, as  will  be  shown. 

As  in  Art.  28,  let  r  =  length  of  crank,  I  =  length  of  connecting  rod,  c  =  r/l, 
n  =  number  of  revolutions  of  crank  per  unit  time  (assumed  constant),  co  = 
angle  in  radians  described  by  crank  per  unit  time  (co  =  2  ttw),  s  =  the  varying 
distance  of  the  crosshead  from  its  position  most  remote  from  the  crank,  d  = 
the  crank  angle  PdOP,  and  t  =  time  required  for  the  crank  to  describe  the 
angle  6  {Q  =  oit  =  2  -wnt).  It  follows  from  the  geometry  of  the  figure,  as 
explained  in  Art.  28,  that 

s=  {l-\.r)  -l{i-  c"^  sin2  6)^  -  r  cos  6. 

Now  (i  —  c^  sin^  0)2  =  I  —  I  c^  sin^  6  —  \d^  sin^  6  —  etc.  (binomial  expansion). 
And  since  c  is  generally  less  than  \,  the  third  and  succeeding  terms  in  the  series 
are  very  small  and  negligible;  hence  we  have  as  a  good  approximation 

Ui  -  c2  sin2  0)^  =  /  (i  -  i  c2  sin^  0)  =  /  (i  -  1  c^  +  1  c^  cos  2  d), 

and  5  =  r  (i  —  COS0)  +  4  cr  (i  —  cos  2  0).  (i) 

Now  if  we  differentiate  this  with  respect  to  t,  we  get  ds/dt  or  v  (velocity  of  the 
crosshead),  and  remembering  that  dd/dt  =  co,  we  finally  get 

V  =  r(ji  (sin  0  +  ^ c  sin  26).  (2) 

Differentiating  again  and  remembering  that  w  is  constant  we  get  dv/dt  or 

a  =  ru)^  (cos  6  -{-  c  cos  2  6).  (3) 

Because  of  our  way  of  measuring  s,  the  positive  direction  is  from  the  cylinder 
toward  the  crank.  Positive  velocity  v  means  that  the  crosshead  is  moving 
toward  the  crank,  and  positive  acceleration  a  means  that  velocity  toward  the 
crank  is  being  added  to  the  velocity. 

In  order  to  furnish  a  comparison  between  the  foregoing  approximate  formu- 
las and  the  exact  ones  of  Art.  28,  we  give  in  the  adjoining  table  the  values  of  a 
for  the  case  c  =  32  for  a  few  values  of  the  crank  angle  d  (Fig.  243). 


e 

a,  exact 

a,  approx. 

0° 

+  1.286  rw2 

+  1.286  r«2 

30 

60 

90 

120 

+  IO15 

+0.357 
—  0.298 
-0.643 

+  1.009 

+0.357 
-0.286 
-0.643 

ISO 
180 

-0.717 
-0.714 

-0.723 
-0.714 

To  find  the  acceleration  of  the  crosshead  when  the  crank  is  at  the  "  head-end 
dead-center  "  (crank  at  OPo)  we  put  6  =  o,  and  find  from  either  the  exact  or 
approximate  formula  that 

a  =  rco^  (i  -\-  c). 


Art.  30 


137 


Length  of  Rod 
Length  of  Crank 


0—0- 
O  1 


m  w  V 

Rod  —o-  equals  -o- J  Cranho- 
»  — o—  „  -0-4  i>  -o — 
t>    -o—     »?    -0-5   "    ~° — 

;;     -o—       V      -o-  Q     „    -o 

»- J. //.//. -5 ^ 


■yi    VUVlll 
— o — 0-0 


-0-0 
-0-0 


7  8 


Fig.  244 


To  find  the  acceleration  at  the  "crank-end  dead-center"  we  put  6  =  180°,  and 
find  from  either  the  exact  or  approximate  formula  that 

a  =  —  rw-  (i  —  c). 

To  show  that  the  motion  of  the  crosshead  C  is  approximately  simple  har- 
monic we  show  that  its  motion  resembles  the  motion  of  Q  (Fig.  243)  which  is  a 
simple  harmonic  one.  In  Fig.  244  we  have  marked  nine  corresponding  posi- 
tions of  Q  and  C.  Thus  points  o  to 
8  are  the  positions  of  Q  when  the  crank 
angles  are  0°,  225°,  45°,  etc.,  and 
points  O,  I,  II,  III,  etc.,  are  the  corre- 
sponding positions  of  C.  In  the  lower 
part  of  Fig.  244  the  paths  of  Q  and  C 
(with  the  points  i,  2,  3  and  I,  II,  III, 
marked  upon  them)  have  been  brought 
together  for  comparison.  It  is  seen 
that  the  actual  distances  described  by 
Q  and  C  in  any  interval  of  time  are 
nearly  the  same,  and  so  the  motion 
of  C  is  nearly  the  same  as  that  of  Q. 
The  three  intermediate  lines  in  the  figure  are  paths  of  C  with  points  corre- 
sponding to  I,  2,  3,  etc.,  for  three  other  lengths  of  connecting  rod.  And  we 
see  that  the  longer  the  rod  the  more  nearly  is  the  motion  of  the  crosshead 
simply  harmonic. 

To  arrive  at  a  more  complete  comparison  of  the  motions  of  C  and  Q,  we  will 
derive  the  formulas  for  the  position,  velocity,  and  acceleration  of  Q  correspond- 
ing to  equations  (i),  (2),  and  (3).     The  variable  distance  of  Q  from  Po  (Fig. 

243)  we  will  call  z,  then 

s  =  r  (i  —  cos  6).  (4) 

Differentiating  with  respect  to  /,  we  get  for  velocity  of  Q 

V  =  rcc  sin  6,  (5) 

and  differentiating  again  we  get  for  accel- 
eration of  Q 

a  =  ro}"^  cos  9.  (6) 

Now  compare  (i)  and  (4),  (2)  and  (5),  and 
(3)  and  (6)  and  note  that  the  formulas  for 
the  motion  of  C  contain  an  "extra  "  term. 
Each  of  these  terms  depends  on  c  (=  r/l), 
or  on  the  "obliquity  "  of  the  connecting 
rod  (maximum  inclination  of  the  rod  to  the 
line  of  stroke  OC).  The  smaller  c  (the 
longer  the  rod  in  comparison  with  the 
crank),  the  smaller  are  the  extra  terms,  and  so  the  longer  the  rod  the  more 
nearly  is  the  motion  of  the  crosshead  a  simple  harmonic  one. 


Fig.  245 


138  Chap,  vii 

Fig.  245  presents  a  comparison  of  the  motion  of  the  crosshead  C  and  the 
motion  of  Q.  The  solid  lines  refer  to  the  first  motion  and  the  dashed  lines  to 
the  second.  Vc  is  the  velocity-distance  {v-s)  graph  and  Ac  is  the  acceleration- 
distance  (as)  graph  for  the  motion  of  C.  Vq  is  the  velocity-distance  graph 
and  Aq  is  the  acceleration-distance  graph  for  the  motion  of  Q.  The  graphs 
for  C  were  drawn  for  a  connecting  rod  three  cranks  long  (c  =  i  -^  3).  For 
longer  rods  the  graphs  for  C  would  come  much  nearer  the  graphs  for  Q. 

31.   Motion  and  Force 

The  preceding  discussion  of  motion  deals,  for  the  most  part,  with  displace- 
ment, velocity,  and  acceleration;  it  does  not  refer  at  all  to  the  forces  acting 
upon  the  moving  bodies.  In  this  article  we  explain  in  what  manner  any  rec- 
tilinear motion  of  a  rigid  body  depends  upon  the  forces  acting  upon  it. 

§  I.  First  View  and  Form  of  the  Fundamental  Principle.  —  In  Art. 
2  it  is  explained  that  the  units  of  force  most  used  by  engineers  are  the  so-called 
gravitation  iinits,  equal  to  the  earth-pulls  on  certain  things  called  standards  of 
weight.  These  units  have  slightly  different  values  at  different  places;  thus 
we  have  the  London  pound-force,  the  New  York  pound-force,  etc.  Some 
writers  define  the  pound-force  as  any  force  equal  to  the  earth's  attraction  on 
the  standard  pound  weight  at  London  or  at  sea  level  in  latitude  45°,  thus 
making  the  unit  force  invariable  or  an  "absolute  "  one.  Besides  these  units 
there  are  others;  see  §  2  of  this  article. 

In  Art.  2,  we  explained  also  that  the  word  weight  is  used  in  at  least  two 
senses  in  common  parlance  (see  footnote,  page  4).  But  we  will  continue  to 
use  it  in  a  single  sense,  to  connote  the  earth-pull  on  a  body,  and  we  employ  a 
separate  word  (mass,  see  §  2  of  this  article)  to  connote  the  amount  of  substance 
or  stuff  in  a  body.  Our  two  weighing  devices,  beam-scale  and  spring-scale, 
differ  in  a  certain  feature  which  is  worth  noting  here.  A  beam-scale  measures 
the  weight  (earth-pull)  of  a  body  in  terms  of  the  local  unit  of  force,  say  the 
pound  force  for  the  place  where  the  weighing  is  done;  a  spring-scale  measures 
the  weight  of  a  body  in  terms  of  an  invariable  unit,  say  the  particular  pound 
force  for  which  the  scale  was  graduated.  A  beam-scale  will  not  detect  the 
change  in  the  weight  of  a  body  with  change  of  place  because  the  magnitude  of 
the  unit  (pull  on  the  poise)  changes  just  as  the  weight  of  the  body  changes. 
A  spring-scale  if  sufficiently  accurate  will  detect  change  in  weight  with  change 
of  place. 

First-hand  knowledge  of  the  relation  between  motion  and  the  forces  acting 
on  the  moving  body  must  rest  on  observation  or  experiment.  Let  us  consider 
a  simple  case  of  motion,  that  of  a  falling  body.  The  motion  takes  place  under 
the  action  of  the  weight  of  the  body  and  the  resistance  of  the  surrounding  air. 
But  if  the  falling  body  is  quite  dense,  the  air  resistance  is  negligible  compared 
to  the  weight  until  the  velocity  becomes  quite  large.  Observations  have  shown 
that  such  a  body  falls  with  a  constant  acceleration  of  about  32  feet  per  second 
per  second  at  moderate  velocities,  and  we  infer  that  any  force  equal  to  the 


Art.  31  139 

weight  of  the  body  would,  if  acting  alone  on  that  body,  produce  an  acceleration 
of  the  value  stated. 

We  are  now  led  to  inquire  what  is  the  effect  on  a  body  of  an  applied  force  of 
some  other  magnitude,  say  a  force  equal  to  double  its  weight  or  one-half  its 
weight?  If  we  could  intensify  or  dilute  the  earth-pull  upon  a  body  by  a 
(gravity)  lens  or  screen,  then  we  could  make  a  body  fall  under  a  force  differing 
from  its  own  weight  and  ascertain  the  answer  to  our  question  by  observing  the 
fall.  Unfortunately  for  our  purpose,  we  cannot  so  concentrate  or  dilute  the 
force  of  gravity  but  we  can  dilute  it  indirectly  by  means  of  an  "Atwood  ma- 
chine," designed  for  that  purpose.  The  essential  parts  of  that  machine  are  a 
light  pulley  P  mounted  on  a  smooth  horizontal  axle  (Fig.  246),  some  blocks  of 
metal  which  can  be  suspended  as  shown  by  a  light  flexible 
cord,  and  a  timing  device  for  getting  the  acceleration  of  A 
and  B  when  the  system  is  allowed  to  move.  Neglecting  the 
small  influence  of  the  pulley,  axle,  and  cord,  we  regard  A  and 
B  as  the  body  moved  and  the  difference  in  their  weights 
(Wb  —  Wa)  as  the  driving  force.  Experiments  with  this  ma- 
chine show  that  A  and  B  move  with  constant  acceleration, 
and  when  runs  are  made  with  various  driving  forces  —  all 
metal  pieces  being  used  each  time  —  then  the  acceleraUons  in  the  ^^^  ., 
differeni  runs  are  directly  proportional  to  the  driving  forces.  In 
this  machine  the  driving  force  can  be  made  very  small  but  it  cannot  be  made 
larger  than  the  weight  of  all  the  metal  pieces.  It  would  seem  that  the  force- 
acceleration  relation  stated  holds  even  for  driving  forces  larger  than  the  weight 
of  the  body  moved;  and  we  assume  that  when  any  forces  are  applied  successively 
to  the  same  body  so  as  to  make  it  move  in  g,  straight  line,  then  the  accelerations  are 
proportional  to  the  forces  respectively.  Or,  if  F  and  F'  =  the  magnitudes  of  two 
forces  applied  to  any  body  in  succession,  and  a  and  a'  =  the  accelerations 
respectively,  then 

F/F'  =  a/a'. 

If  IF  =  the  weight  of  the  body,  g  =  the  acceleration  due  to  gravity  (IF),  F 

and  a  as  above,  then  the  foregoing  principle  gives  also  F/W  =  a/g,  or  as  it  is 

more  commonly  written,  F  =  (W/g)  a. 

Generally,  a  moving  body  is  under  the  influence  of  more  than  one  force. 

When  the  body  moves  in  a  straight  line,  the  resultant  of  all  the  forces  acting  upon 

it  is  a  single  force  acting  in  the  direction  of  the  acceleration  (proved  in  Art.  35). 

Therefore  the  resultant  has  no  component  at  right  angles  to  the  line  of  motion; 

or,  the  algebraic  sum  of  the  components  of  all  the  forces  acting  on  the  body  along 

any  line  at  right  angles  to  the  path  equals  zero.     Thus,  if  the  path  is  taken  as  an 

X  axis  and  two  lines  at  right  angles  to  each  other  and  to  the  path  as  y  and  z 

axes,  then 

SF„=o,    2F,  =  0,     and    ^F,  =  R, 

where  SPx,  "^Fy,  and  llF^  stand  for  the  algebraic  sums  of  the  x,  y,  and  2  com- 


140 


Chap,  vii 


ponents  of  all  the  forces  acting  on  the  body,  and  R  denotes  their  resultant. 

Furthermore,  as  proved  in  Art.  35, 

W 

(i) 


J?      ^^ 
K  =  — a. 


Any  unit  of  force  may  be  used  for  R  and  W  in  equation  (i),  and  any  unit  for 
g  and  a.  When  a  gravitational  unit  of  force  is  used  —  such  are  most  conven- 
ient in  engineering  calculations  —  then,  strictly,  the  numerical  value  of  g  used 
should  correspond  to  the  "locality  "  of  the  unit-force  used.  That  is,  when  one 
is  about  to  make  a  calculation  by  means  of  equation  (i),  implying  the  New 
York  pound-force  say,  then  he  should  use  for  g  its  value  for  New  York.  As 
already  stated,  the  variation  in  g  is  negligible  in  most  engineering  calculations, 
and  we  generally  use  32.2  feet  per  second  per  second  or  even  32  for  simplicity. 
Non-gravitational  units,  the  dyne  for  example,  may  be  used  in  equation  (i). 
But  when  such  units  are  preferred,  then  equation  (2)  is  to  be  preferred  in  place 
of  equation  (i). 

Examples.  —  When  a  body  moves  in  a  straight  line  and  if  all  the  forces  act- 
ing on  it  are  known  so  that  R  can  be  computed,  then  the  acceleration  can  be 
determined  easily  by  means  of  equation  (i).  If  the  acceleration  is  known 
then  we  can  determine  R  easily,  and  from  R  we  can  find  out  something 
about  the  forces  acting  on  the  body. 

I.  A  (Fig.  247)  represents  a  body  being  dragged  along  a  rough  horizontal 
surface  5  by  a  pull  P  acting  as  shown.  Suppose  that  the  body  weighs  100 
pounds,  P  =  40  pounds,  and  the  friction  resistance  =  10  pounds.  We  will 
find  the  acceleration  of  A  and  the  normal  component  of  the  force  exerted 
between  A  and  B.  The  forces  acting  on  A  are  represented  in  Fig.  248,  N  de- 
noting the  normal  component  of  the  reaction  oi  B  on  A,  friction  being  the  other 
component.  Resolving  at  right  angles  to  the  path,  we  get  N  +  40  sin  20° 
=  100,  or  iV  =  86.3  pounds.  Resolving  along  the  path,  we  get  i?  =  40  cos  20° 
—  10  =  27.6.  Equation  (i)  gives  27.6  =  (100  -r-  32.2)  a,  or  a  —  8.9  feet  per 
second  per  second. 


A    _- 


too  I  /b3. 


Fig.  247 


1^ 


/o 


Fig.  248 


40, 
lbs. 


Fig.  249 


Fig.  250 


2.  A  (Fig.  249)  represents  a  body  being  dragged  up  the  rough  inclined  plane 
5  by  a  pull  P  equal  to  50  pounds;  A  weighs  60  pounds  and  the  coefficient  of 
friction  for  A  and  B  is  I.  We  determine  the  acceleration.  Three  forces  act 
on  A,  namely  the  weight,  the  pull,  and  the  reaction  of  B.  The  last  force  is 
represented  by  two  components  (TV  and  F)  in  Fig.  250.  Resolving  at  right 
angles  to  the  path,  we  get  N  =  60  cos  30°  =  52 ;  hence  F  =  ^2  -^  4=  1^  pounds. 
Resolving  along  the  path,  we  get  7?  =  50  —  13  —  60  sin  30°  =  7  pounds;  hence 
7  =  (60  -^  32.2)  a,  or  a  =  3.75  feet  per  second  per  second. 


Art.  31 


141 


3.  A  certain  passenger  elevator  gets  up  speed  at  the  rate  of  4  feet  per  second 
per  second,  and  can  be  stopped  at  the  rate  of  8  feet  per  second  per  second. 
We  discuss  the  pressure  on  the  shoes  of  a  standing  passenger  weighing  160 
pounds,  during  an  ascent.  The  forces  acting  on  the  man  are  his  own  weight 
and  the  pressure  P  of  the  floor  on  his  shoes  (upward).  During  acceleration  the 
resultant  of  these  forces  is  upward,  hence  P  is  larger  than  160  pounds  and 
R  =  P  —  160.  Equation  (i)  becomes  P  —  160  =  (160  -f-  32)  X  4  =  20,  or 
P  —  180  pounds.  During  the  next  period,  constant  speed,  a  =  o  and  P  —  160. 
During  retardation  the  acceleration  is  downward  and  hence  R  gilso.  There- 
fore R  =  160  —  P  =  (160  -h  32)  X  8  =  40,  or  P  =  120  pounds. 

4.  We  determine  the  reaction  of  the  car  (Fig.  251)  on  A  during  the  period 
of  getting  up  speed  at  the  rate  of  2  feet  per  second  per  second;  A  weighs  1000 
pounds.  We  suppose  the  floor  of  the  car  so  rough  that  A  does  not  slip.  There 
are  two  forces  acting  on  A  (Fig.  252),  its  own  weight  and  the  pressure  P  of 
the  floor.  This  latter  force  must  be  inclined  as  shown  to  furnish  a  component 
on  A  in  the  direction  of  the  acceleration.  Resolving  at  right  angles  to  the 
path,  we  get  P  cos  6  =  1000;  resolving  along  the  path,  we  get  R  =  P  sind  = 
(1000  -^  32.2)  X  2.  Solving  these  two  simultaneously  we  find  that  P  =  1002 
pounds  and  ^  =  3°  Zi' •  (The  horizontal  component  of  P  is  friction.  To 
prevent  slipping  the  floor  must  be  rough  enough  to  furnish  such  a  resistance.) 


■lOOOIbi 


^3-1 


Fig.  251 


^6'>i. 


Fig.  252 


Fig.  253 


150 
Ibi, 

^ 

/so  lbs. 

/        N\ 

^p^^ 

Fig. 

254 

Fig. 

2SS 

5.  A  box  (Fig.  253)  containing  a  body  A  slides  down  a  rough  inclined  plane  B 
whose  inclination  is  40°.  The  box  weighs  300  pounds,  A  weighs  150  pounds, 
the  coefficient  of  kinetic  friction  "between  "  box  and  plane  is  J,  and  A  is  per- 
fectly smooth.  We  determine  the  acceleration  of  A  and  box,  and  the  pressures 
between  them.  Fig.  254  represents  all  external  forces  acting  on  A  and  box,  the 
reaction  of  the  plane  B  being  represented  by  two  components,  F  and  N.  Re- 
solving at  right  angles  to  the  path,  we  get  N  =  150  cos  40°  +  300  cos  40° 
=  345,  hence  F  =  345  -f-  5  =  69.  Resolving  along  the  path,  we  get  R  =  300 
sin  40°  +  150  sin  40°  —  69  =  220  ^  (450  -^  32.2)  a,  or  a  =  15.75  feet  per 
second  per  second.  Fig.  255  represents  all  the  forces  acting  on  A ,  where  P  and 
Q  are  the  pressures  exerted  by  the  front  and  bottom  of  the  box  respectively. 
Resolving  along  the  direction  of  the  motion  we  get  7?  =  150  sin  40°  —  P 
=  (150-7-32.2)15.75,  or  P  =  23.1.  Resolving  along  the  normal  we  get 
()  =  150  cos  40°  =115  pounds. 

6.  A  body  slides  down  a  plane  under  the  influence  of  gravity  and  the  re- 
action of  the  plane  only;  required  the  acceleration.  Let  W  =  weight  of  the 
body;   fi  =  coefficient  of  friction;  a  =  inclination  of  the  plane;  N  —  normal 


142  Chap,  vii 

pressure;  and  F  =  friction.  Then  resolving  forces  normally  to  the  path  we 
get  iV  =  M^  cos  a;  therefore  F  =  ijlN  =  /AV  cos  a.  Resolving  along  the  path 
we  get  R  =  W  sina  —  F  =  W  (sin  a  —  /jl  cos  a)  =  {W  -^  g)  a,  or 

a  =  g  (sin  a  —  (J.  cos  a) . 

If  the  plane  is  perfectly  smooth  n  =  o,  and  a  —  g  sin  a. 

§  2.  Second  View  and  Form  of  the  Fundamental  Principle.  —  Physi- 
cists avoid  the  (common)  double  meaning  of  the  word  weight  by  employing 
the  word  mass  to  connote  amount  of  material,  substance,  or  stuff,  in  a  body, 
and  weight  to  connote  the  earth-pull  on  the  body.  Such  usage  is  followed  in 
this  book.  Material  is  measured  in  different  ways;  for  example,  Hquids  gen- 
erally by  gallon,  earthwork  by  cubic  yard,  cloth  by  square  yard,  brick  by  thou- 
sand, iron  by  ton,  etc.  But  mass  means  amount  of  substance  as  measured 
by  a  beam-scale.  Our  standards  of  mass  (commonly  and  legally  called  "  stand- 
ards of  weight  ")  are  the  pound  and  the  kilogram.  These  are  certain  pieces 
of  metal  preserved  in  London  and  Paris  respectively.  The  mass  of  a  body, 
measured  as  just  explained,  does  not  change  with  change  of  locality,  and  this 
is  in  accordance  with  our  conception  of  material,  substance,  or  stuff. 

The  force-acceleration  relation,  F  =  (W/g)  a,  can  be  put  into  an  alternative 

form  which  is  preferable  from  some  points  of  view.     Thus  suppose  that  two 

bodies  whose  weights  at  the  same  place  are  Wi  and  W2  are  subjected  to  equal 

forces  F;   let  g  =  the  acceleration  due  to  gravity  at  the  place  and  ai  and  a^ 

=  the   accelerations  produced   by  the   two   forces  F.     Then  F  =  (Wi/g)  Ci 

=  Wi/g)  02,  or 

ai/a2  =  W2/W1. 

That  is,  the  accelerations  of  the  two  bodies  are  inversely  as  their  weights  at  the 
same  place;  and  since  the  masses  of  two  bodies  are  proportional  to  the 
weights  (at  the  same  place),  the  accelerations  of  the  two  bodies  are  inversely 
proportional  to  their  masses.  This  relation  and  that  between  the  accelerations 
produced  in  a  body  by  two  different  forces  acting  singly  can  be  expressed  in 
one  statement  as  follows:  —  Whenever  a  force  acts  upon  a  body  so  as  to  make  it 
move  in  a  straight  line,  then  the  acceleration  produced  is  proportional  to  the  force 
directly  and  to  the  mass  of  the  body  inversely,  or  a  ^F  -^  m.  This  proportion- 
ality can  be  put  into  the  form  of  an  equation, 

F  =  Kma, 

where  i^T  is  a  proportionality  factor  whose  value  depends  on  the  units  used 
for  expressing  magnitudes  of  F,  m,  and  a.  This  is  the  alternative  form 
mentioned. 

We  may  fix  the  value  of  K  in  two  ways:  —  (i)  choose  units  of  F,  m,  and  a 
at  pleasure,  and  deduce  the  value  of  X;  or  (2)  choose  a  value  of  K  and  units  for 
any  two  of  the  quantities  F,  m,  and  a,  and  then  deduce  the  proper  unit  for  the 
third  quantity.  On  plan  (i)  we  choose,  for  example,  the  pound-force,  the 
pound-mass,  and  the  foot  per  second  per  second  as  units  for  F,  m,  and  a,  and 


Art.  31  143 

then  determine  K  by  reference  to  any  motion  in  which  F,  m,  and  a  are  known. 
The  motion  of  a  falling  body  is  such  a  one.  Thus  when  a  body  "weighing  " 
say  10  pounds  falls,  then  F  =  10  pounds,  m  =  10  pounds,  and  a  =  about  32.2 
feet  per  second  per  second,  and  we  have  10  =  X"  X  10  X  32-2,  or  K  =  i  -i- 
32.2.  On  plan  (2)  we  take  K  equal  to  unity  for  simplicity,  and  then  (i)  choose 
units  of  m  and  a  at  pleasure,  and  deduce  the  proper  unit  of  F;  or  (ii)  choose 
units  of  F  and  a  at  pleasure,  and  deduce  the  proper  unit  of  m.  (i)  Physicists 
take  the  gram  as  unit  mass,  and  the  centimeter  per  second  per  second  as  unit 
of  acceleration;  then  the  corresponding  unit  of  force  {K  =  i)  is  such  a  force  as 
would  give  to  the  gram  an  acceleration  of  one  centimeter  per  second  per 
second.  They  call  this  force  the  dyne,  (ii)  If  we  take  the  pound  as  unit  of 
force,  the  foot  per  second  per  second  as  unit  of  acceleration,  then  the  corre- 
sponding unit  of  mass  (K  =  i)  is  such  a  mass  which  will  sustain  an  acceleration 
of  one  foot  per  second  per  second  under  the  action  of  a  force  of  one  pound. 
This  unit  of  mass  has  no  generally  accepted  name,  but  it  is  sometimes  called 
"engineers'  unit  of  mass,"  also  "slug  "  and  "gee-pound." 

A  set  of  units  for  which  A'  =  i  is  called  a  systematic  set  of  units,  also  a 
kinetic  set.  We  will  always  use  systematic  units  and  thus  always  have  F  =  ma, 
or  when  several  forces  make  a  body  move  in  a  straight  line, 

R  =  ma.  (2) 

where  R  denotes  the  resultant  of  those  forces.  For  a  falling  body  R  =  W  and 
a  =  g;  thus  when  systematic  units  are  used 

W  =  mg,    or    m  =  W/g.  (3) 

Therefore  R  =  (W/g)  a  as  in  §  i. 

To  arrive  at  a  notion  of  the  magnitude  of  the  unfamiliar  units  dyne  (force) 
and  slug  (mass),  let  us  consider  the  well-known  force-mass-acceleration  rela- 
tion in  the  case  of  a  falling  body.  A  body  whose  mass  is  one  gram,  falling  at 
Paris,  falls  under  the  action  of  a  force  (earth-pull)  of  one  Paris  gram,  and  has 
an  acceleration  of  981  centimeters  per  second  per  second.  Hence  a  force  of 
0.001019  (=  I  -H  981)  Paris  grams  would  give  to  a  body  whose  mass  is  one 
gram  an  acceleration  of  one  centimeter  per  second  per  second.  Therefore 
that  force  is  the  dyne,  that  is 

I  dyne  =  0.001019  Paris  grams  (force). 

A  body  whose  mass  is  one  pound,  falling  at  London,  falls  under  the  action  of  a 
force  (earth-pull)  of  one  London  pound,  and  has  an  acceleration  of  32.2  feet 
per  second  per  second.  Hence  a  force  of  one  London  pound  would  give  to  a 
body  whose  mass  is  32.2  pounds  an  acceleration  of  one  foot  per  second  per 
second.    Therefore,  that  mass  is  the  slug,  that  is 

.   I  slug  =  32.2  pounds  (mass). 


CHAPTER   VIII 

CURVILINEAR  MOTION 
32.  Velocity  and  Acceleration 

§  I.  Velocity.  —  In  common  parlance,  velocity  of  a  moving  point  at  a 
certain  instant  means  the  rate  at  which  the  point  is  describing  distance  then. 
So  understood,  velocity  has  magnitude  and  sign  only,  and  is  therefore  a  scalar 
quantity.  In  the  preceding  chapter  (on  rectilinear  motion)  we  used  the  word 
in  this  sense;  in  the  present  chapter  we  use  the  word  in  a  broader  sense  —  so 
that  it  is  a  vector  quantity  whose  magnitude  is  the  rate  at  which  the  moving 
point  is  describing  distance  at  the  instant  in  question  and  whose  direction  is 
the  same  as  that  of  the  motion  then. 

If  5  =  the  (varying)  distance  of  the  moving  point  from  some  fixed  origin 
in  the  path,  the  distance  being  measured  along  the  path,  then  the  magnitude 
of  the  velocity  at  any  instant  equals  the  value  of  ds/dt  for  that  instant.  Or  if 
V  =  magnitude  of  velocity, 

V  =  ds/dt.  (i) 

If  the  point  is  moving  uniformly,  then  the  rate  at  which  distance  is  described 
is  constant,  and  is  given  by  As/^t,  where  As  is  the  distance  described  in  any 
interval  At.  The  direction  of  the  motion  at  any  instant  (and  the  direction  of 
the  velocity  then)  is  along  the  tangent  to  the  path  at  the  position  of  the 
moving  point  at  that  instant.  To  illustrate,  imagine  a  lo-foot  wheel  mounted 
on  a  horizontal  axis  which  points  north  and  south,  and  suppose  that  the  wheel 
is  rotating  at  180  revolutions  per  minute  clockwise  when  viewed  from  the 
south.  When  a  certain  point  on  the  rim  is  in  its  highest  position  then  the 
velocity  of  the  point  has  a  magnitude  of  2  tt  5  X  180  =  5655  feet  per  minute, 
and  the  direction  of  the  velocity  is  horizontal  from  west  to  east. 

The  magnitude  part  of  a  velocity  is  called  speed  by  some  writers;  we  follow 
this  usage.  Thus  in  the  preceding  illustration  the  speed  is  5655  feet  per  minute; 
while  the  wheel  turns,  the  speed  of  the  point  is  constant  but  the  velocity 
changes  in  direction. 

§  2.  Acceleration.  — The  acceleration  of  a  moving  point  at  any  instant 
is  the  rate  at  which  its  velocity  —  not  speed  —  is  changing  then.  If  V  de- 
notes the  (varying)  velocity  of  a  moving  point  and  v  the  (varying)  speed,  then 
the  definition  states  that  the  acceleration  is  dV/dt  and  not  dv/dt.  Inasmuch 
as  most  readers  are  unfamiliar  with  the  rate  of  a  vector  quantity  —  the  rate 
chapters  in  most  books  on  differential  calculus  deal  with  rates  of  scalar  quanti- 
ties only  —  we  explain  in  considerable  detail  just  what  is  meant  by  the  rate  ot 

144 


Art.  32 


H5 


change  of  a  velocity,  but  first  we  explain  for  subsequent  use  a  motion  graph 
called 

Hodo graph.  —  This  is  a  curve  which  shows  how  the  velocity  of  a  moving 
point  varies.  It  is  constructed  by  laying  off  vectors  from  a  point  to  represent 
successive  velocities,  and  then  the  free  ends  of  the  vectors  are  joined  by  a 
smooth  curve.  The  curve  is  the  hodograph  for  the  motion.  Thus,  suppose 
that  A  BCD  (Fig.  256 )  is  the  path  of  a  moving  point  P,  and  that  the  vectors  at 
A,  B,C,  and  D  represent  the  velocities  of  P  when  at  A,  B,  C,  and  D  respec- 
tively.   If  0'A\  O'B',  O'C,  and  O'D'  (Fig.  257)  are  drawn  (from  any  point  O') 


Fig.   256 


a       bed 
Fig.  258 


to  represent  the  velocities  respectively,  then  the  curve  A'B'C'D'  is  the  hodo- 
graph for  the  motion  of  P  from  ^  to  D.  The  increment  or  change  in  the 
velocity  while  P  moves  from  /I  to  D  say  is  represented  by  the  vector  A'D' 
(in  magnitude  and  direction).  The  change  in  the  speed  =  length  O'D'  — 
length  O'A'.  (The  hodograph  should  not  be  confused  with  the  speed- time 
curve.  The  latter  is  represented  in  Fig.  25S  where  ab,  he,  and  cd  represent  the 
times  required  for  P  to  move  from  A  to  B,  B  to  C,  and  C  to  D  respectively,  and 
the  ordinates  over  a,  b,  c,  and  d  represent  the  speeds  2XA,B,  C,  and  D.) 


Fig.  259 


Fig.  260 


1.8    2,0     2.2    2.4Secs. 
Fig.  261 


We  are  now  ready  to  explain  the  meaning  of  rate  of  change  of  velocity;  we 
base  our  explanation  on  a  simple  case  of  curvilinear  motion.  Suppose  that  a 
point  P  starts  at  Q  (Fig.  259)  and  describes  the  circle  shown  in  such  a  way  that 
the  distance  traversed  (in  feet)  equals  double  the  cube  of  the  time  after  start- 
ing (in  seconds),  or  5  =  2  t^.  Required  the  acceleration  say,  when  t  =  2.4 
seconds,  or  5  =  2  X  2.4^  =  27.65  feet.  The  curve  in  Fig.  260  is  the  hodograph 
of  the  motion  for  the  interval  from  /  =  1.6  to  i  =  2.6,  containing  the  instant 


146 


Chap,  viii 


in  question.     It  was  constructed  from  the  adjoining  schedule,  computed  from 
s  =  2t^,d  =  s/20  (radians)  =  2.865  ^  (degrees),  and  v  =  ds/dt  =  6  P. 


t  (sec.) 

s  (ft.) 

e  (deg.) 

V  (ft/sec) 

1.6 

8.192 

23-5 

15-36 

1.8 

I I . 664 

33-4 

19.44 

2.0 

16.000 

45-8 

24.00 

2.2 

21 . 296 

61.0 

29.04 

2.4 

27.648 

79.2 

34-56 

2.6 

35-152 

100.  7 

40.56 

Vectors  O'A',  O'B',  O'C ,  etc.,  represent  the  velocities  of  F  when  /  =  1.6,  1.8, 
2 .0,  etc. ,  as  marked.  Vectors  A  'E' ,B'E',  C'E' ,  and  D'E'  represent  the  velocity- 
increments  for  the  intervals  1.6  to  2.4,  1.8  to  2.4,  2.0  to  2.4,  and  2.2  to  2.4 
seconds  respectively.  The  magnitudes  of  these  increments  were  scaled  from 
the  original  hodograph  drawing  (the  scale  of  which  was  one  inch  =  5  feet  per 
second)  and  are  recorded  in  the  adjoining  schedule  under  A  V. 


At  (sec.) 

AV  (ft/sec) 

AV/At 
(ft/sec/sec) 

Av  (ft/sec) 

Av/At 
(ft /sec /sec) 

1.6  to  2.4  =  0.8 
1.8  to  2.4  =  0.6 
2.0  to  2.4  =  0.4 
2.2   to  2.4  =  0.2 

28.75 
25-15 
19-55 
11-45 

35-9 
41.9 

48-9 

57-2 

19.20 
15.12 
10.56 

5-52 

24.0 
25.2 
26.4 
27.6 

Now  the  magnitude  of  the  average  acceleration  for  the  interval  1.6  to  2.4 
seconds  is  28.75  "^  ^-^  =  35-9  feet  per  second  per  second,  and  the  direction 
of  that  average  acceleration  is  A'E'.  The  magnitudes  of  the  average  accelera- 
tions for  the  intervals  1.8  to  2.4,  2.0  to  2.4  and  2.2  to  2.4  also  are  given  under 
AF/A/;  the  directions  of  those  average  accelerations  are  respectively  B'E', 
C'E',  and  D'E'. 

Now  the  acceleration  when  /  =  2.4  seconds  is  the  limit  of  these  average 
accelerations,  as  At  is  taken  smaller  and  smaller  but  always  terminating  at 
/  =  2.4.  The  magnitude  of  this  limit,  which  is  the  magnitude  of  the  accelera- 
tion sought,  is  the  limit  of  the  magnitudes  of  the  average  accelerations.  This 
limit  can  be  found  approximately  by  plotting  as  in  Fig.  261,  where  ordinates 
equal  to  the  computed  average  accelerations  were  erected  at  the  proper  points 
on  the  time-base,  thus  determining  the  solid  curve  abed.  Any  other  ordinate 
represents  an  average  acceleration  for  an  interval  terminating  at  /  =  2.4 
seconds,  thus  the  ordinate  at  2.1  represents  the  average  acceleration  for  the 
interval  2.1  to  2.4.  The  curve  abed  may  be  extended  a  short  distance  without 
error,  and  therefore  the  ordinate  over  /  =  2.4  is  approximately  the  limit  of  the 
values  35.9,  41. Q,  etc.,  and  it  represents  closely  the  magnitude  of  the  accelera- 
tion at  ^  =  2.4  seconds.    This  ordinate  scales  66.5  feet  per  second  per  second 


Art.  32  147 

on  the  original  drawing  already  mentioned.  The  direction  of  this  acceleration 
is  the  limit  of  the  directions  of  the  average  acceleration,  and  obviously  this 
limit  is  the  tangent  to  the  hodograph  at  £'.  On  the  original  drawing  the  angle 
between  this  tangent  and  the  horizontal  is  24  degrees. 

For  emphasis  by  contrast  we  will  determine  the  way  in  which  the  speed 
changes  during  the  motion  under  consideration.  Speed-increments  are  listed 
under  H-o  in  the  schedule;  average  rates  of  change  of  speed  for  the  respective 
time-intervals  are  listed  under  A17  A/.  The  luniting  value  of  these  averages, 
as  A^  is  taken  smaller  and  smaller  but  always  termmating  at  /  =  2.4,  is  about 
28  feet  per  second  per  second,  and  this  is  the  rate  at  which  the  speed  changes 
(dv/dt)  at  /  =  2.4  seconds. 

We  now  generalize  the  foregoing.  Let  A B  (Fig.  262)  be  the  path  of  a  moving 
point  P,  and  let  O'A'  and  O'B'  be  the  velocities  of  P  when  at  .4  and  B  respec- 
tively. Then  vector  A'B'  is  the  velocity-in- 
crement for  the  interval  A/  while  P  moves 
from  A  to  B;  (chord  A'B')  -^  A/  is  the  mag- 
nitude of  the  average  acceleration  for  the  in- 
terval, and  the  direction  A'B'  is  the  direction 
of  the  average  acceleration.  The  magnitude 
of  the  (instantaneous)  acceleration  of  P  when  ^^^-  ^^^ 

passing  A  is  the  limit  of  (chord  .1  'B')  4-  A/,  as  B  is  taken  closer  and  closer  to 
A]  and  the  direction  of  the  acceleration  is  the  limit  of  the  direction  of  A'B' 
as  B  approaches  A,or  B'  approaches  A'.  Now  hm  (chord  A'B')  -^  A^  =  lim 
(arc  A'B')  -i-  A/  =  ds' /dt  where  ds'  is  the  elementary  portion  of  the  hodo- 
graph at  A',  and  s'  is  the  distance  of  P'  (the  point  in  the  hodograph  corre- 
sponding to  P)  from  any  fixed  origin  on  the  hodograph;  and  the  limiting 
direction  of  the  chord  A'B'  is  the  tangent  at  A'.     Finally, 

the  acceleration  of  P  is  a  vector  quantity  equal  to  ds' /dt  and  parallel 
to  the  tangent  to  the  hodograph  at  the  point  P'  corresponding  to  P. 
It  should  be  noted  that  the  acceleration  of  P  is  not  directed  along  the  tan- 
gent to  the  path  but  always  toward  the  concave  side  of  the  path.     It  may  be 
noted  also  that  since  the  velocity  of  P'  equals  ds'/dt  and  is  directed  along  the 
tangent  to  the  hodograph  at  P', 

the  acceleration  of  P  is  the  same  as  the  velocity  of  its  rorresponding  point  P', 

it  being  understood  that  s'  (distance  on  the  hodograph)  must  be  measured 
by  the  scale  of  the  hodograph  diagram. 

As  an  example  of  the  use  of  our  final  result,  that  the  acceleration  of  P  is  given 
by  the  velocity  of  its  corresponding  point  in  the  hodograph,  we  determine  the 
acceleration  of  a  point  which  describes  a  circle  at  a  constant  speed.  Let  P 
(Fig.  263)  be  the  point,  r  =  radius  of  the  circle,  and  v  =  the  speed  of  P.  The 
hodograph  is  a  circle  whose  radius  equals  v\  A'  corresponds  to  A  and  P'  to  P; 
and  hence  A  'O'P'  equals  6.  We  measure  the  distance  5  (traversed  by  P)  from 
A,  and  s'  (traversed  by  P')  from  A'.     Then  s'/v  =  s/r,  or  s'  =  sv/r.     Now 


148 


Chap,  vin 


the  velocity  of  P'  equals  ds' /dt  =  (ds/dt)  (v/r)  =  v'^/r,  and  the  velocity  of  P' 
is  directed  along  the  tangent  at  P'  (parallel  to  the  radius  OP);    hence  the 

acceleration  of  P  is  directed  from  P  to  O  and 
its  magnitude  is  v^/r. 

The  method  for  determining  acceleration 
used  in  the  preceding  example  is  diflScult  to 
apply  in  most  motions.  Why  then  was  the 
method  developed  at  length?  To  make  plain 
the  meaning  of  acceleration  in  curvilinear 
motion  and  particularly  to  show  students, 
in  an  elementary  way,  that  acceleration  in 
curvilinear  motion  does  not  equal  dv/dt  and  is  not  directed  along  the  tan- 
gent to  the  path  in  general.  Thus  in  the  preceding  example  it  was  found 
that  the  magnitude  of  the  acceleration  is  v^/r,  whereas  dv/dt  =  o  since  v 
is  constant;  also  it  was  found  that  the  acceleration  is  directed  along  the 
normal  to  the  path.  In  the  motion  discussed  at  length  (where  5  =  2  /^),  it 
was  found  that  the  magnitude  of  the  acceleration  when  /  =  2.4  seconds  is 
about  66.5  feet  per  second  per  second;  but,  smce  v  =  ds/dt  =  6  f,  dv/dt  = 
12  t  =  28.8  for  t  =  2.4.* 


Fig.  263 


33.   Components  of  Velocity  and  Acceleration 

§  I.  Components  of  Velocity.  —  Velocity,  like  any  other  vector  quantity, 
can  be  resolved  into  components.  For  our  purpose  components  parallel  to  axes 
of  coordinates  (as  x,  y,  and  s)  are  most  useful;    such  components  are  called 

*  Note  on  Rate  of  Change  of  a  Vector  Quantity.  —  We  shall  have  to  deal  with  the  rates  of 
vector  quantities  other  than  velocity.  Therefore  we  now  generalize  our  notions  on  the  rate 
of  this  vector  quantity  (velocit}')  just  arrived  at  so  as  to  prepare  for  the  rates  of  these  other 
vector  quantities  for  future  use.  Let  OA,  OB,  OC,  etc.  (Fig.  264),  represent  successive 
values  of  any  vector  p,  in  magnitude  and  direction,  vector  OB  represent- 
ing p  at  time  ti,  OB  at  time  to,  OC  at  time  /s,  etc.  The  changes  in  p  during 
the  intervals  ti  to  t^,  ti  to  tz,  /i  to  t^,  etc.,  are  represented  by  the  vectors 
AB,  AC,  AD,  etc.  The  average  rate  of  change  in  the  vector  p  during 
any  of  these  intervals  may  be  found  by  dividing  the  change  by  the 
time;  thus  for  the  interval  /i  to  tt  the  average  rate  =  AB  -r-  (^2  —  /i), 
and  this  rate  is  a  vector  whose  direction  is  AB.  For  the  interval  t\  to 
tz,  the  average  rate  =  AC  -i-  {tz  —  /i)  and  the  direction  of  the  rate  is  AC. 
In  general,  both  the  magnitude  and  the  direction  of  the  average  rate 
of  a  vector  depend  on  the  length  of  the  interval  for  which  the  average 
rate  is  taken  or  computed.  By  true  or  instantaneous  rate  of  change  of 
the  vector  at  the  time  ti,  say,  is  meant  the  limit  of  the  average  rate  AB  -^  {t2  —  /i)  as  ^2  is  taken 
closer  and  closer  to  ti.  The  magnitude  of  this  limit  =  limit  of  chord  AB  -h  (t-i  —  ti)  =  limit 
of  arc  AB  -r-  {t2  —  /i)  =  dS/dt  where  dS  =  elementary  portion  of  the  arc  ;  the  direction  of 
the  hmit  is  the  direction  of  the  tangent  to  the  arc  at  A. 

Imagine  a  point  P  to  move  in  the  curve  AD  so  that  the  vector  OP  represents  the  vector 
p  at  each  instant.  The  velocity  of  P  =  dS/dt  and  its  direction  at  any  instant  is  tangent  to 
the  curve  at  the  point  where  P  is  at  the  instant;  hence  the  time-rate  of  p  is  the  same  as  the 
velocity  of  P  (the  moving  end  of  p). 


Fig.  264 


Art.  ss 


149 


Axial  Components. — Let  x,  y,  and  s  =  the  (changing)  coordinates  of  a 
moving  point  P,  and  v^,  Vy,  and  v^  =  the  components  of  the  velocity  of  P 
parallel  to  the  x,  y,  and  s  coordinate  axes  respectively;   then 

•i'l  =  dx/dt,        Vy  =  dy/dt,         v^  =  dz/dt.  (i) 

(Proof  follows.)  These  formulas  state  that  each  axial  component  of  the 
velocity  at  any  instant  equals  the  rate  at  which  the  corresponding  coordinate 
of  the  moving  point  is  changing  then.  In  the  following  derivation  of  the  for- 
mulas we  assume  for  simplicity  that  the  path  of  the  moving  point  is  a  plane 
curve  —  in  the  xy  plane;  proof  can  be  extended  readily  to  include  the  case  of 
a  tortuous,  or  twisted,  path.  Let  P  (Fig.  265)  be  the  moving  point,  v  =  the 
magnitude  of  the  velocity  of  P,  and  a  =  the  angle  which  the  tangent  at  P 
makes  with  the  x  axis.  Then  v^  =  v  cos  a,  and  Vy  =  v  sin  a.  But  v  =  ds/dt, 
cos  a.  =  dx/ds,  and  sin  a  =  dy/ds;  hence 

_  ds  dx  _  dx  ,        _dsdv  _dy 

"'"dlTs  ~Tt'   ^"^"^    '"'~  dtds  ~  dl' 


Fig.  265 


Fig.  266 


For  an  example,  we  determine  the  x  and  y  components  of  the  velocity  of  a 
point  P  which  moves  in  the  circle  of  Fig.  266  according  to  the  law  5=2  t'^,  s 
being  in  feet  and  /  in  seconds.  (This  is  the  motion  discussed  at  length  in  the 
preceding  article.)  It  is  plain  from  the  figure  that  x  =  20  cos  0  =  20  cos 
(5/20)  =  20  cos  (o.i^^);  hence 

i)x  =  —  20  sin  (o.i/^)  0.3 1-  =  —  6f  sin  {p.if). 

When  /  =  2  seconds,  say,  zj^  =  —  6  X  4  sin  (0.8  radians)  =  —  24  sin  45.8°  =  — 
17.2  feet  per  second.  The  negative  sign  means  that  the  component  of  the 
velocity  is  directed  toward  the  left.  In  a  similar  way  it  can  be  shown  that 
Vy=  6/^  cos  {o.if). 

Other  Components.  —  The  velocity  of  a  moving  point  P  is  directed  along  the 
tangent  to  its  path  at  the  point  where  P  is  at  the  instant  under  consideration; 
hence,  the  tangential  component  of  the  velocity  equals  the  velocity  itself,  and 
the  velocity  has  no  normal  component  (along  the  normal  to  the  path).  For 
formulas  for  components  of  velocity  along  and  perpendicular  to  the  radius- 
vector  of  the  moving  point  see  Hoskins'  "Theoretical  Mechanics,"  Ziwet's, 
or  any  other  standard  work  on  that  subject. 

§  2.  Components  of  Acceleration. — Acceleration  is  a  vector  quantity,  and 
can  be  resolved  into  components  therefore.     The  most  useful  components  for 


I50 


Chap,  vtri 


our  purposes  are:  —  (i)  Those  parallel  to  axes  of  coordinates  {x,  y,  and  z),  called 
axial  components;  (2)  those  parallel  to  the  tangent  and  normal  to  the  path  of 
the  moving  point  at  the  place  where  the  point  is  at  the  instant  in  question.* 
Axial  Components. — Let  a^,  ay,  and  az  =  the  axial  components  of  the 
acceleration  of  a  moving  point  P,  and  as  in  §  i  let  Vx,  Vy,  and  Vz  =  the  (varying) 
axial  components  of  the  velocity  of  P,  then 

flx  =  dVx/dt,         Oy  =  dvy/dl,         a^  =  dvjdt.  (2) 

(Two  proofs  follow.)  These  formulas  state  that  each  axial  component  of  the 
acceleration  of  P  at  any  instant  equals  the  rate  at  which  the  corresponding 
axial  component  of  the  velocity  of  P  is  changing  then.  Since  Vx  =  dx/dt, 
Vy  =  dy/dt,  and  v^  =  dz/dt,  we  have  also 

ax  =  d^x/df,        ay  =  d'-y/df^,        a,  =  dh/df^.  (3) 

In  the  following  proof  it  is  assumed  for  simplicity,  that  the  path  of  the 

moving  point  is  a  plane  curve ^ —  in  the  xy  plane.     The  proof  can  be  extended 

readily  to  include  the  case  of  a  tortuous  or  twisted  path.     LetP  (Fig.  267)  be 

the  moving  point,  and  O'P^  (Fig.  268)  be  parallel  and  equal  to  the  velocity  v\ 


Fig.  268 


Fig.  267 

then  P'  is  the  point  "corresponding  "  to  P,  and  the  direction  of  the  accelera- 
tion of  P  is  tangent  to  the  hodograph  at  P'  as  indicated.  Let  a  =  the  mag- 
nitude of  the  acceleration,  and  a'  =  the  angle  between  the  acceleration  and 
the  X  axis.  Then  ax  =  a  cos  a  and  ay  =  a  sin  a'.  But  a  =  ds'/dt,  where  ds' 
denotes  elementary  length  on  the  hodograph  (see  Art.  32);  and  since  the 
coordinates  of  P'  are  Vx  and  Vy,  cos  a'  =  dvx/ds',  and  sin  a'  =  dvy/ds'.     Hence 

_  ds'  dvx  _  dvx  ,  _  ^  ^  _  ^  4. 

^'~l[tdl'~dt'     ^"""^     ''''  ~  dt   ds'  ~  dt  • ' 


y 


*  For  discussion  of  components  along  and  perpen- 
dicular to  the  radius-vector  drawn  from  any  fixed 
origin  to  the  moving  point  see  texts  referred  to  in  §  i . 
t  The  following  is  an  alternative  proof:  — Let  AB 
(Fig.  269)  be  a  portion  of  the  path  of  the  moving  point 
P,  and  let  O'A'  and  O'B'  represent  the  velocities  of  P 
^     '^  when  at  A  and  B.     Then  A'B'  represents  the  change 

^^^'  ^9  jj^  ^Yic  velocity  while  P  moves  from  A  to  B,  and  A'M 

and  A'N  represent  the  x  and  y  components  of  this  velocity-change.     Let  A'Q,  tangent  to  the 
hodograph  at  ^',  represent  the  acceleration  of  P  when  at  A.     Then 


1     N      , 

\B' 

1      / 

/-A 

1   . 

A  1  i 

// 

/  \'/ 

1  "■ 

iM 

1 

ax 


a  cos  a 


A'B'        ,         ,.,\r.      ,•     A'B'cosB'A'M      ..     A'M 
hm  ^  lim  (cos  B'A'M)  =  hm rr =  hm 


But  A'M  =  O'X  -  O'X'  =  increment   in   the  x  component  of   the  velocity  =  Avx] 
«as  =«:  lim  (Avi/At)  =  dVx/dt.    In  a  similar  way  one  could  prove  that  a„=  dvy/dt. 


hence 


Art.  ,-i$ 


151 


For  an  example  we  determine  the  x  and  y  components  of  the  acceleration 
in  the  motion  of  the  preceding  example  (see  Fig.  266).  In  that  example  it  was 
shown  that  the  general  value  of  the  x  component  of  the  velocity  (true  for  any 
instant)  is  Vx  =  —  6  /-  sin  (o.i  t^) ;  hence 

dvx/dt  (or  Qx)  =  —  12  t  sin  (o.i  i^)  —  1.8  /^  cos  (o.i  l^). 

And  when  /  =  2.4  seconds,  say,  Gx  —  —  29.4  feet  per  second  per  second.  In  a 
similar  way  the  value  of  Cy  can  be  found  from  the  general  expresssion  for  Vy. 

Tangential  and  Normal  Componenls. — They  will  be  denoted  by  at  and  a„ 
respectively;  other  notation  as  before,  and  r  =  radius  of  curvature  of  the  path 
at  the  point  occupied  by  the  moving  point  at  the  instant  in  question.     Then 

at  =  dv/dt  =  d'^s/dt",        and        c„  =  v'^/r.  (4) 

(Two  proofs  follow.)  These  formulas  respectively  state  that  at  =  the  rate 
at  which  the  speed  (magnitude  of  the  velocity)  changes,  and  that  a„  is  propor- 
tional to  the  square  of  the  speed  directly  and  to  the  radius  of  curvature  inversely. 
Where  the  speed  is  increasing,  dv/dt  is  positive  and  at  has  the  same  direction 
as  the  velocity;  where  the  speed  is  decreasing,  dv/dt  is  negative  and  at  is 
opposite  to  the  velocity  in  direction.  The  normal  acceleration  a„  is  always 
directed  from  the  mo\ang  point  toward  the  center  of  curvature.  (The  words 
tangential  and  normal  refer  to  the  tangent  and  normal  to  the  path  at  the 
point  where  the  moving  point  is  at  the  instant  in  question.) 


0 

Fig.  270  Fig.  271 

Let  AB  (Fig.  270)  be  the  path  of  a  moving  point  P,v  =  velocity  of  P  at  ^, 
V  -\-  ^v  =  its  velocity  at  B,  and  Ad  =  the  angle  between  the  normals  (and  the 
tangents)  to  the  path  at  A  and  B.  Also  let  O'A'  and  O'B'  be  equal  to  and  par- 
allel to  V  and  v  +  Av  respectively;  then  A'  and  B'  are  on  the  hodograph  and 
angle  A'O'B'  =  Ad.  The  acceleration  of  P  when  at  A  is  parallel  to  the  tan- 
gent A'Q.  Let  A'Q  represent  c;  then  A'AI  and  A'N  respectively  represent 
the  tangential  and  normal  components  of  a.     Hence 


at  =  acos(f)  =  -77  cos  <^, 
dt 


and     a„  =  a  sin  4> 


To  continue  the  proof,  we  need  to  recall  certain  formulas  from  calculus.  Let 
CC  (Fig.  271)  be  any  curve,  O  a  convenient  "pole,"  p  and  p  +  Ap  the  radius 
vectors  of  C  and  C,  A7  the  angle  COC,  Al  the  arc  CC,  yp  the  angle  between 
OC  and  the  tangent  at  C.     From  calculus, 

siiwp  =  p  dy/dl     and     cos  1/'  =  dp/dl. 


152 


Chap,  vin 


These  formulas  when  applied  to  the  hodograph  (Fig.  270)  become 

sm4>  —  V  dd/ds'    and    cos  4>  =  dv/ds'. 
Hence 

ds'  dv       dv  ,  ds'     dd         dB        dd  ds      v^  ^ 

dt  ds        dt  dt    ds  dt        ds  dt       r 

For  an  example  we  determine  the  tangential  and  normal  components  of  the 
acceleration  in  the  motion  of  the  two  preceding  examples.  Since  5=2^^, 
V  =  6  t"  and  dv/dt  =  12/  =  a«;  at  /  =  2.4  seconds,  say,  at  =  28.8  feet  per 
second  per  second.  Also  a„  =  v'^/r  =  36  ^^,20  =  1.8  /^;  at  /  =  2.4,  a„  =  59.7 
feet  per  second  per  second. f 

The  {resultant)  acceleration  can  be  obtained  from  its  axial  or  tangential  and 
normal  components.     Thus 

a  =  V  a/  +  Cy^  +  flz'-  =  Vat^  +  an\ 
The  angles  which  a  makes  with  the  x,  y,  and  z  axes  are  given  respectively  by 

cos~^  (a^c/a),  cos~^  {ay/ a),  and  cos~^  {az/a). 
The  angle  which  a  makes  with  the  normal  equals  tan"^  {at/ an).  From  a  = 
{at^  +  a„^)5  it  appears  that  a  does  not  equal  at  {=  dv/dt)  in  general;  onJ^  when 
an  =  o.  And  an{=  "v^/r)  =  o  only  when  v  =  o  or  r  ^  oc,  that  is,  where  the 
moving  point  reverses  direction  of  motion  or  where  the  radius  of  curvature  is 
infinitely  great. 

*  The  following  is  an  alternative  proof:  —  Let  AB  (Fig.  272)  be  a  portion  of  the  path  of 
the  moving  point  P,  and  O'A'  and  O'B'  represent  the  velocities  of  P  when  at  A  and  B  respec- 
tively. Then  the  curve  A'B'  is  the  hodograph  for  .IB;  the  chord  A'B'  represents  the  change 
in  the  velocity  while  P  moves  from  A  to  B;  and  the  tangent  A' a  represents  the  accelera- 
tion a  of  P  when  at  A.  Let  v  =  the  magnitude  of  the  velocity  a,t  A,  v  -\-  Av  =  that  at  B, 
Ad  —  the  angle  between  the  normals  (and  the  tangents)  at  A  and  B,  and  0  the  angle  be- 
tween the  acceleration  and  the  velocity  at  A.     Then  at  =  a  cos(^  = 

,.     A'B',.     ,       ^,,„„       ,.     A'B' coiEA'B'      ,.     A'E      ,.      (ii -}- Ai.)  cos  Ai*  -  » 
hm lim  (cos,  EAB)  =  lim =  nm  — —  =  lim — = 

A^  Ai  At  At 

,.     AvcosA6  —  v(i  —  cosA9)       ,.     Av        .„  ,.      sin-|A9      dv       i    ,.     Ad  ^^      dv 

hm -r =  lim  --  cos  A0  —  2V  hm  — — —  =  -j. ti  hm  —  A9  =  -y 

At  At  At  dt       2  At  dt 

Referring  to  the  figure  it  will  be  seen  that  a„  =  a  sin  0  = 

,.     A'B'..      ..    ^.,^,.       ,.     A'B'^mEA'B'      ..     EB'  (j^±Av)^nA9 

hm-— —  lim  (smEA'B')  =  hm- — =  hm— -—  =  hm — = 

At  At  At  At 

,.     sinAff   ,    ,.     Av   .     ,„         ,.    Ad  ,  dO        \  ds      -fi 

V  hm  — h  lim  -—  sin  Ad  =  v  hm— -  +  o  =  v-j-=  "»-  -7;  =  —  • 

At  At  At  dt        r  dt      r 

t  Some  students  find  it  diflficult  or  impossible  to  realize  that  acceleration  of  a  point  in 
curvilinear  motion  has  a  component  along  the  normal  to  the  path  at  the  place  where  the 
moving  point  is  at  the  instant  in  question,  notwithstanding  detailed  calculations  (as  on  pages 
145  and  146)  for  a  specific  case  and  mathematical  derivation  of  the  general  formulas  for  the 
normal  component  of  acceleration. 

Let  us  consider  the  matter  from  the  perplexed  student's  own  standpoint.  He  may  ask, 
"  If  the  moving  point  has  an  acceleration  along  the  normal,  why  does  it  not  acquire  velocity 
along  the  normal  ?  "  If  he  will  grant  that  velocity  cannot  be  acquired  instantaneously  but 
only  with  lapse  of  time,  then  it  is  easy  to  show  that  velocity  is  acquired  along  the  normal. 
Thus  to  take  a  concrete  case,  suppose  that  a  point  P  is  moving  in  the  curve  (Fig.  272);  con- 


ART.  33 


153 


Simple  Harmonic  Motion  (see  Art.  30). — The  fact  that  the  components 
of  the  velocity  and  acceleration  —  along  any  line  —  of  a  moving  point  P 
equal  the  velocity  and  acceleration  of  the  projection  of  the  point  on  that  same 
line,  enables  one  to  get  the  formulas  for  velocity  and  acceleration  in  a 
simple  harmonic  motion  very  easily.  Thus  let  P,  Fig.  273,  be  a  point  describ- 
ing the  circle  uniformly,  and  Q  its  projection  on  the  horizontal  diameter; 
then  the  motion  of  ()  is  a  simple  harmonic  one  (Art.  30).  Let  the  amplitude  of 
the  s.h.m.  (radius  of  the  circle)  =  2  feet,  and  the  frequency  of  the  s.h.m. 
(revolutions  of  P  per  unit  time)  =  100  vibrations  per  minute.  Then  the 
velocity  ofP=27rX2Xioo  =  i26o  feet  per  minute  =21  feet  per  second, 
directed  along  the  tangent  at  P  as  shown;  and  the  acceleration  of  P  =  21-  -^  2 
=  220  feet  per  second  per  second,  directed  along  the  radius  PO.  Now  when 
PO  makes  an  angle  d  =  30°  say,  then  the  velocity  of  Q  is  21  sin  30°  =  10.5 
feet  per  second;  the  acceleration  oiQ  =  220  cos  30°  =  iSo  feet  per  second  per 
second,  directed  toward  O  whether  P  is  travelling  clockwise  or  counter  clock- 
wise. Evidently  the  greatest  velocity  of  Q  obtains  when  ()  is  at  0;  that  value 
equals  21  feet  per  second.  The  greatest  acceleration  of  Q  obtains  when  Q  is  at 
either  end  of  its  path;  that  value  is  220  feet  per  second  per  second. 


Fig.  273  Fig.  274 

General  formulas  for  velocity  and  acceleration  in  simple  harmonic  motion 
can  be  as  easily  derived.  Let  r  =  amplitude,  n  =  frequency.  Then  the 
velocity  of  P  =  2  irrn  and  its  acceleration  =  4  ir-r-n^  H-  r  =  4  Trhi^r.  Hence 
velocity  and  acceleration  of  Q  are  respectively  (see  Fig.  273) 
—  2  irrn  sin  d  and  —  4  ir'^n-r  cos  6. 
§  3.  Projectile  Without  Air  Resistance.  —  Let  u  =  the  velocity  of 
projection  (initial  velocity),  and  a  =  the  angle  of  projection  (angle  between 
direction  of  projection  and  the  horizontal),  x  and  y  =  the  coordinates  of  the 
projectile  P  (Fig.  274)  at  any  time  t  after  projection,  v  =  the  velocity  of  P,  and 

sider  the  interval  while  P  moves  from  A  to  B.  Let  O'A'  and  O'B'  represent  the  velocities 
at  A  and  B  respectively;  then  A'B'  represents  the  velocity  acquired  by  P  in  the  interval, 
and  this  acquired  velocity  has  a  component  not  only  along  the  normals  at  A  and  B,  but 
along  any  other  normal  to  AB.  Or,  the  student  may  say,  "Since  the  velocity  is  always 
directed  along  the  tangent  —  and  hence  has  no  normal  component  —  there  cannot  be  an  accel- 
eration along  the  normal."  But  here  is  a  case  which  may  convince  him:  a  ball  thrown 
obliquely  into  the  air.  The  acceleration  of  the  ball,  due  to  gravity,  is  at  all  times  vertically 
down,  and  this  acceleration  has  for  every  position  of  the  ball  a  component  along  the  normal 
at  that  position.  (Strictly  the  acceleration  is  not  quite  vertical  by  reason  of  air  resistance, 
but  neglecting  this  fact  is  of  no  consequence  here.) 


154  Chap,  vni 

a  =  the  acceleration  of  P  at  the  time  /.  The  only  force  acting  on  the  pro- 
jectile during  flight  is  gravity.  Hence  the  acceleration  of  the  projectile  is 
vertically  downwards  at  all  times  and  equal  to  g  (Art.  34),  ox  a^  =  o  and 
dy  =  —  g.  Since  there  is  no  x  acceleration,  the  x  velocity  remains  constant 
during  the  flight,  and  we  find  that  value  from  the  initial  conditions  (m,  a)  to  be 

Vx  =  ucosa.  (i) 

The  y  velocity  is  decreased  at  all  times  by  the  y  acceleration  —  g.  In  the 
interval  /,  that  decrease  is  gt,  and  since  the  initial  y  velocity  is  u  sin  a,  the  v 
velocity  at  any  time  /  is  given  by 

Vy  =  u  sin  a  —  gt.  (2) 

Since  the  x  velocity  remains  constant,  the  x  displacement  in  the  interval  /  is 

given  by 

X  =  u  cos  a  '  t.  (3) 

The  y  velocity  varies  uniformly  with  the  time;  hence  the  average  y  velocity 
for  the  interval  /  is  |  [(w  sin  a  +  {u  sin  a  —  gt)]  =  w  sin  a  —  |  gt.  The  y  dis- 
placement for  the  interval  equals  the  product  of  the  average  y  velocity  and 

the  time  or 

y  =  usina  •  t  —  ^  gf^.  (4) 

Foregoing  results  determine  the  velocity  and  position  at  any  time  /.  They 
may  be  arrived  at  more  directly  by  integrating  the  given  equations 

dvx  ,  dvu 

Thus  integrating  the  first  equation  we  find  that  Vx  =  C\,  where  Ci  is  a  constant 
of  integration  whose  value  for  reasons  already  stated  is  u  cos  a.  Integrating 
the  second  equation  we  find  that  Vy  =  —  gt  -{-  C2  where  C2  is  another  constant 
of  integration.  From  the  initial  conditions  Vy  =  m  sin  a  when  /  =  o,  and  on 
substituting  these  values  of  Vy  and  t  in  the  last  equation  we  find  that  C2  = 
M  sin  a;  thus  Vy  =  —  gt  -\-  u  sin  a  as  before.  Now  integrating  Vx  =  dx/dt  = 
u  cos  a,  we  get  x  =  u  cos  a  •  /  -f  C3.  From  initial  conditions  x  =  o  when 
/  =  o;  therefore  v  =  o  -\-  Cz,  or  C3  =  o,  and  x  =  u  cos  a  •  ^  as  before.  Inte- 
grating Vy  =  dy/dt  =—  gt  -]r  usm.  a,  we  get  y  =  —  \  gf^  -{-  usina- 1  -\-  d. 
From  initial  conditions  y  =  o  when  /  =  o;  therefore  o  =  o  -{-  o  -|-  C4  or 
Ci  =  o,  and  y  =  —  \  gt^  -\-  u  sina  '  t  as  before. 

The  trajectory  (path  of  the  projectile)  is  a  portion  of  a  parabola  as  can  be 
shown  from  the  equation  of  the  trajectory.  To  arrive  at  the  equation  we  may 
combine  equations  (3)  and  (4)  so  as  to  eliminate  /.     Thus  we  find  that 

y  2u'^  cos^  a  =  xu^  sin  2  a  —  gx^.  (5) 

Range  aftd  Greatest  Height.  —  At  the  end  X  of  the  range,  y  =  o;  hence  the 
time  of  flight  is  given  by  u  sin  at  —  ^  gt^  =  o,  or  t  =  (2  w  sin  a)/g.  The  range 
R  equals  the  value  of  x  in  equation  (3)  when  /  =  the  value  just  found;  thus 

R  =  (u^  sin  2  a)  -7-  g. 


Art.  34  155 

R  also  equals  the  value  of  x  in  equation  (5)  when  y  =  o.     The  formula  for  R 

shows  that  the  range  is  greatest  —  for  a  given  velocity  of  projection  —  when 

a  =  45°.     That  greatest  value  is  u'^/g. 

At  the  highest  point  of  the  trajectory  Vy  =  o;   hence  the  time  of  flight  to 

that  point  is  given  by  m  sin  a  —  g/  =  o,  or  /  =  {u  sin  a)  -^  g.     The  height  H 

of  the  trajectory  equals  the  value  of  y  in  equation  (4)  when  t  =  the  value  just 

found ;   thus 

H  =  ^  {usin  a)2  -^  g. 

H  also  equals  the  value  of  y  in  equation  (5)  when  x  =  ^  R. 

34.   Motion  of  the  Center  of  Gravity  of  a  Body 

In  Art.  3 1  we  found  that  any  rectilinear  motion  of  a  body  depends  in  a  very 
simple  way  upon  the  forces  acting  on  the  body.  The  relation  between  the 
motion  of  the  center  of  gravity  of  a  body  (whether  rigid  or  not)  which  has  any 
sort  of  motion  however  complicated  is  also  quite  simply  related  to  the  forces 
exerted  on  the  body  as  we  shall  see  presently. 

§  I.  A  Particle  is  a  body  so  small  that  its  dimensions  are  negligible  in 
comparison  with  the  range  of  its  motion.  In  any  motion  of  a  particle  no  dis- 
tinction need  be  made  between  the  displacements  (velocities  or  accelerations) 
of  different  points  of  the  particle,  for  they  are  equal,  or  practically  so;  and 
by  displacement  (velocity  or  acceleration)  of  the  particle  is  meant  the  dis- 
placement (velocity  or  acceleration)  of  any  point  of  the  particle. 

"Laws  of  Motion.'"  —  i.  When  no  force  is  exerted  upon  a  particle  then  it 
remains  at  rest  or  continues  to  move  uniformly  in  a  straight  line.  2.  When  a 
single  force  is  exerted  upon  a  particle^  then  it  is  accelerated;  the  direction  of  the 
acceleration  is  the  same  as  the  direction  of  the  force,  and  its  magnitude  is  propor- 
tional to  the  force  directly  and  to  the  mass  of  the  particle  inversely.  3.  When  one 
particle  exerts  a  force  upon  another,  then  the  latter  exerts  one  on  the  former;  and 
the  two  forces  are  equal,  colinear,  and  opposite. 

These  are  essentially  Newton'' s  Laws  of  Motion.  The  form  of  statement  here 
used  differs  however  from  that  in  which  he  announced  them  (1687).  They 
are  based  on  observation  and  experience.  Newton  was  led  to  them  through 
his  study  of  the  motions  of  heavenly  bodies.  No  other  moving  bodies  have 
been  so  accurately  and  extensively  observed,  and  the  agreement  of  the  laws 
and  those  motions  constitutes  the  best  evidence  of  the  correctness  of  the  laws. 

Law  3  has  already  been  referred  to  (page  43,  footnote).  This  law  is  doubted 
by  some  beginners  in  this  subject.  The  doubt  is  sometimes  expressed  in  this 
way:  "When  a  horse  pulls  on  a  cart,  then,  if  the  cart  pulls  back  on  the  horse 
an  equal  amount  (as  the  law  states),  why  is  it  that  they  generally  move  for- 
ward? "  Close  attention  to  the  forces  which  act  on  the  horse  and  on  the  cart 
should  clear  up  this  doubt.  There  are  three  forces  exerted  on  the  horse,  — • 
his  weight  (exerted  by  the  earth),  the  pull  of  the  cart,  and  the  reaction  exerted 
by  the  roadway  on  his  hoofs.     When  the  horizontal  (forward)  component  of 


iS6 


Chap,  viii 


the  reaction  on  his  hoofs  exceeds  the  pull  back  by  the  cart  then  the  horse  starts 
forward.  There  are  three  forces  acting  on  the  cart,  — •  its  weight  (exerted 
by  the  earth),  the  pull  of  the  horse,  and  the  reaction  of  the  roadway  on  the 
wheels.  When  the  pull  exceeds  the  horizontal  (backward)  reaction  of  the 
roadway  then  the  cart  starts  forward.  Or,  the  motion  of  horse  and  cart 
together  may  be  explained  like  this:  There  are  four  forces  acting  upon  the 
pair,  —  the  weight  of  the  horse,  that  of  the  cart,  the  reaction  of  the  roadway  on 
the  horse,  and  that  on  the  cart;  the  horse  and  cart  start  to  move  when  the 
horizontal  component  of  the  reaction  of  the  roadway  on  the  horse  exceeds  that 
on  the  cart. 

Law  2  is  discussed  at  length  in  Art.  3 1  for  the  case  of  rectilinear  motion,  but 
is  not  referred  to  there  as  a  "law."  It  covers  curvilinear  motion,  as  well  as 
rectilinear,  inasmuch  as  no  reference  to  kind  of  motion  is  made  in  the  law. 
We  cannot  give  a  real  illustration  of  a  particle  moving  under  the  action  of  a 
single  force.  But  miagine  a  particle  projected  in  some  way,  and  then  sub- 
jected to  a  single  force  inclined  to  the  direction  of  projection;  the  particle 
would  move  in  a  curved  path.  (A  ball  in  flight  through  the  air  is  a  near  ap- 
proach to  our  imagined  illustration.  This  ball  is  acted  upon  by  two  forces, 
earth-pull  and  air  resistance;  at  moderate  velocities  the  latter  may  be  neg- 
ligible in  comparison  with  the  former.)  The  law  states  that  the  direction  of 
the  acceleration  of  the  particle  agrees  at  each  instant  with  the  direction  of  the 
force,  and  that  the  magnitude  of  the  acceleration  is  directly  proportional  to 
the  force  and  inversely  proportional  to  the  mass  of  the  particle  (a  <x  F  ^  m, 
where  a  =  the  acceleration,  m  —  the  mass  of  the  particle,  and  F  =  the  force 
acting  upon  it).  It  is  shown  in  Art.  31,  §  2,  that  the  proportion  a  cc  F/ni  can 
be  written  as  an  equation  F  =  Kma  where  i»C  is  a  constant  whose  value  depends 
on  the  units  used  for  F,  m,  and  a.  Units  may  be  chosen  so  that  K  =  1;  such 
units  are  "systematic  units";  for  example,  dyne  (force),  gram  (mass),  and 
centimeter  per  second  per  second  (acceleration).  We  will  continue  to  take 
iiC  =  I  (as  in  Art.  31),  thus  implying  the  use  of  systematic  units. 

Law  I  is  really  included  in  law  2.  For  if  there  is  no  force  acting  on  a  particle 
during  any  particular  interval  of  time,  then  the  particle  has  no  acceleration 
during  the  interval  (according  to  law  2) ;  and  hence  the  velocity  of  the  particle, 
whatever  it  may  be,  remains  unchanged.  Thus,  if  the  velocity  is  zero  at  the 
beginning  of  the  interval,  then  the  velocity  remains  zero,  that  is  the  particle 
rests;  if  the  velocity  is  not  zero  initially,  then  the  velocity  remains  constant 
in  magnitude  and  direction,  that  is  the  particle  moves  uniformly  and  in  a 
straight  line.  This  fact  is  important  enough  to  warrant  its  statement  in  a 
separate  law. 

The  word  inertia  is  used  in  Mechanics  to  refer  to  the  property  of  the  matter 
involved  more  or  less  in  laws  i  and  2.  It  refers  to  the  fact  that  the  natural 
state  of  a  particle  is  rest  or  uniform  rectilinear  motion,  that  a  particle  is  reluc- 
tant, as  it  were,  to  change  that  state,  and  responds  only  to  an  outside  influence 
which  we  call  force.    We  also  express  this  fact  by  saying  that  matter  is  inert. 


Art.  34  157 

"Force  of  inertia  "  is  a  term  which  students  sometimes  use  to  express  a  notion, 
but  generally  in  a  vague  way.  For  example,  concerning  the  motion  of  a 
hockey  puck  projected  without  spin  along  the  surface  of  smooth  ice,  it  is  stated 
sometimes  that  the  puck  is  urged  on  by  the  (or  its)  force  of  inertia.  This 
statement  is  at  variance  with  the  laws  of  motion.  The  only  forces  acting  on 
the  puck,  after  projection,  are  gravity  and  the  reaction  of  the  ice.  There  is  no 
force  urging  the  puck  onward;  it  moves  onward  —  for  a  time  —  because  it 
was  (forcibly)  projected,  and  in  spite  of  the  retarding  influence  of  the  reaction 
of  the  ice.  Were  it  not  for  this  influence  (friction),  the  puck  would  move 
across  the  entire  field  of  ice  at  constant  velocity,  not  because  of  any  force  urg- 
ing it  onward  but  because  of  no  force  to  change  its  natural  state  (of  uniform 
rectilinear  motion). 

For  another  illustration,  imagine  a  yard  stick  mounted  on  a  vertical  axis, 
the  wide  sides  of  the  stick  being  horizontal;  also  imagine  a  coin  laid  on  the 
upper  side  and  near  the  end  of  the  stick  remote  from  the  axis,  and  that  several 
pins  are  stuck  about  the  coin  to  hold  it  in  place  when  the  yard  stick  is  rotated. 
If  the  pins  are  not  too  strong  and  firm,  then  the  stick  may  be  rotated  so  rapidly 
that  the  pins  will  give  way,  and  the  coin  will  "fly  off."  Or,  as  some  would 
say,  the  coin  will  be  "thrown  off  by  the  force  of  inertia."  Such  statement 
is  at  variance  with  the  laws  of  motion.  The  following  is  a  description  of  the 
phenomenon  in  accordance  with  those  laws.  Before  the  stick  is  rotated,  there 
are  two  forces  acting  on  the  coin,  —  its  own  weight  (or  gravity)  and  the  re- 
action of  the  stick  (upward  and  equal  to  the  weight).  When  the  stick  is  ro- 
tated, the  coin  is  forced  into  an  unnatural  state  (curvilinear  motion)  by  some 
of  the  pins.  We  know  from  our  experience  and  observation  that  the  coin 
presses  against  the  outer  pins  (remote  from  the  axis)  and  that  those  pins  press 
against  the  coin.  Thus  there  is  no  force  acting  on  the  coin  tending  to  throw 
it  off  the  stick;  on  the  contrary,  the  pins  exert  forces  to  hold  it  on.  The  coin 
eventually  flies  off  —  as  the  speed  is  increased  —  because  the  pressure  of  the 
coin  against  the  pins  gets  large  enough  to  make  them  give  way;  then  the  pins 
can  no  longer  restrain  the  coin,  and  it  takes  on  a  natural  state  of  motion. 
This  motion  is  along  the  tangent  to  the  (circular)  path  previously  described 
by  the  coin  at  the  point  where  the  coin  is  supposed  to  have 
broken  loose,  and  with  velocity  equal  to  that  of  the  coin  at  failure. 
Of  course  this  natural  motion  is  short-lived,  because  after  the  coin 
has  left  the  stick,  it  is  subjected  to  a  single  unbalanced  force 
(gravity)  which  interferes  with  the  inclination  —  as  it  were  — 
of  the  coin  to  move  along  the  straight  line  mentioned  (tangent). 

When  several  forces  act  on  a  particle  then  the  particle  has  a 
definite  acceleration  at  each  instant,  which  might  of  course  equal 
zero  under  certain  circumstances.  Let  F',  F",  etc.  (Fig.  275),  be  forces  acting 
on  the  particle  P,  and  a  =  the  acceleration,  and  m  =  the  mass  of  P.  Obviously 
some  single  force  R  acting  alone  would  give  the  particle  that  same  accelera- 
tion.    According  to  the  second  law  R  would  have  to  act  in  the  direction  of  the 


iS8 


Chap,  vrn 


acceleration  and  equal  ma.  This  force  is  the  resultant  of  the  forces  F' ,  F",  etc., 
which  actually  produce  the  acceleration.*  Let  a  —  the  angle  between  the  di- 
rection of  the  acceleration  and  any  line,  say  the  x  axes  of  a  coordinate  frame. 
Then  R  cos  a  =  ma  cos  a,  or  Rx  =  max  where  Rx  and  Ox  denote  the  x  com- 
ponents of  R  and  a  respectively.  But  Rx  equals  the  algebraic  sum  of  the  x 
components  of  F',  F",  etc.  (Art.  4),  and  hence  SFj  =  mOx. 

§  2.  Two  OR  More  Particles  considered  collectively  are  called  a  system 
of  particles.  We  conceive  a  body  (whether  rigid  or  not)  as  consisting  of  parti- 
cles, that  is,  as  a  system  or  collection  of  particles.  Among  the  forces  exerted 
upon  any  particle  of  a  body  some  may  be  exerted  by  the  particles  of  another 
body;  such  a  force  has  been  called  an  external  force  with  reference  to  the  body 
under  consideration  (Art.  10).  A  force  exerted  on  a  particle  of  a  body  by 
another  particle  of  the  same  body  is  called  an  internal  force  with  reference  to 
the  body.  According  to  the  third  law  of  motion,  if  one  particle  of  a  body 
exerts  a  force  upon  another,  then  the  second  exerts  a  force  upon  the  first;  and 
these  two  forces  are  equal,  colinear,  and  opposite.  Hence,  a  system  of  internal 
forces  consists  of  pairs  of  equal,  colinear,  and  opposite  forces. 

Let  Fig.  276  represent  a  body,  not  rigid  necessarily,  points  i,  2,  3,  etc.,  being 
constituent  particles  of  the  body;    let  Fi,  F2,  F3,  etc.,  be  the  external  forces 

acting  on  the  body,  the  other  vectors  (not  lettered) 
being  internal  forces.  Imagine  the  last  equation  of  §  i 
(which  states  that  the  algebraic  sum  of  the  com- 
ponents —  along  any  line  —  of  all  the  forces  acting 
on  a  particle  equals  the  product  of  the  mass  of  the 
particle  and  the  component  of  its  acceleration  along 
the  line)  wTitten  down  for  every  particle  of  the  body, 
and  then  imagine  the  left-hand  members  to  be  added 
and  also  the  right-hand  members;  these  sums  are 
equal  of  course.  To  the  first  sum  the  internal  forces 
contribute  nothing,  since  those  forces  occur  in  certain  pairs  as  already  ex- 
plained; hence  the  sum  depends  only  on  the  external  forces.  We  will  de- 
note the  algebraic  sum  of  their  components  along  some  line,  say  an  axis  of  x, 


Fig.  276 


m'a 


X 

/ 


+  m"ax"  + 


where  m', 


by  SFx  as  customarily.     The  second  sum  is 

m",  etc.,  denote  the  masses  of  the  particles  and  Ox',  ax",  etc.,  the  x  components 
of  their  accelerations  respectively.  A  simple  expression  for  this  sum  can  be 
found  as  follows: — Let  x',  x",  etc.,  be  the  ^-coordinates  of  the  particles  at 
any  instant  of  the  motion,  and  x  =  x-coordinate  of  their  mass-center  f  at  that 
instant;  then 


m'x'  +  m"x"  + 


xZjm. 


*  This  foice  R  is  called  resultant  in  accordance  with  the  definition  of  the  term  in  Art.  3, 
where  first  used.  For  if  R  were  reversed,  then  acting  alone  it  would  give  the  particle  an  acceler- 
ation —  a;  and  acting  together  with  the  forces  F',  F",  etc.,  the  acceleration  would  be  zero.  R 
therefore  is  "  equivalent "  to  F',  F",  etc.  All  the  relations  between  concurrent  forces  and  their 
resultant  developed  in  Statics  hold  here  also  for  F',  F",  etc.,  and  R. 

t  Mass-center  is  another  name  for  center  of  gravity.     The  former  term  seems  more  appro- 


Art.  34  159 

Differentiating  with  respect  to  time,  we  get 

v/here  vj ,  vj' ,  etc.,  are  the  x  components  of  the  velocity  of  the  respective  par- 
ticles, and  Vx  is  the  x  component  of  the  velocity  of  the  mass-center.  Differen- 
tiating again  we  get 

m'aj  +  m"ax"  -}-•••=  axSm, 

where  Ux  is  the  x  component  of  the  acceleration  of  the  mass-center.  If  now  we 
equate  these  simplified  expressions  for  the  sums  mentioned  we  get 

2F,  =  Max,  (i) 

where  M  is  written  in  place  of  21  w,  the  mass  of  the  whole  body,  for  simplicity. 

Since 2 Fi  does  not  include  internal  forces,  Cx  does  not  depend  on  those  forces; 
that  is  to  say,  the  acceleration  of  the  mass-center  of  a  system  of  particles  does 
not  depend  at  all  upon  internal  forces. 

Equation  i  is  a  mathematical  form  of  an  important  principle  which  we  will 
call  the  principle  oj  the  motion  of  the  mass-center.  It  may  be  put  into  words  as 
follows:  In  any  motion  of  a  body  {ivhether  rigid  or  not)  the  algebraic  sum  of  the 
components  (along  any  line)  of  all  the  external  forces  equals  the  product  of  the 
mass  of  the  body  and  the  component  of  the  acceleration  of  the  mass-center  along 
that  line.  It  is  worth  noting  that  equation  (i)  is  just  like  the  last  equation  of 
§  I  which  relates  to  the  motion  of  a  particle.  Hence,  the  motion  of  the  mass- 
center  of  a  body  is  the  same  as  though  the  entire  mass  of  the  body  were  con- 
centrated at  the  mass-center  with  all  the  external  forces  acting  on  the  body 
applied  to  such  dense  point  parallel  to  their  actual  lines  of  action.  The  use 
of  systematic  units  (Art.  31)  is  presupposed;  but  if  W/g  be  written  in  place 
of  M  (see  Art.  31,  §  2),  where  W  is  the  weight  of  the  body,  then 

ZFx  =  {W/g)ax  (2) 

and  any  unit  may  be  used  for  F  and  W,  and  any  unit  for  g  and  ax. 

Any  number  of  equations  like  (i)  or  (2)  can  be  written  in  a  given  case  of 
motion,  one  equation  for  each  possible  direction  of  x,  y,  z,  u,  etc.  Only 
three  of  these  equations  would  be  independent;  the  others  would  be  super- 
fluous.    Thus  we  would  have 

SFx  =  Max,        ^Fy  =  May,        2F,  =  Ma,. 

When  the  mass-center  describes  a  curve  then  it  is  usually  more  convenient  to 
resolve  along  the  tangent  to  the  curve,  the  (principal)  normal,  and  at  right 
angles  to  the  plane  of  the  first  two  directions.  The  component  of  the  acceler- 
ation in  this  last  direction  equals  zero;  calling  the  components  in  the  first  two 

priate  in  the  present  discussion.  Since  masses  of  bodies  are  proportional  to  their  weights 
(at  the  same  place),  we  may  substitute  mass  for  weight  in  the  formulas  for  the  coordinates 
of  the  center  of  gravity  (or  mass-center)  in  Art.  21.  (Mass-center  is  generally  defined  with- 
out reference  to  center  of  gravity,  and  then  the  identity  of  the  two  points  is  demonstrated.) 


i6o 


Chap,  vin 


A 

\                           1 

1 

r 

•///// 

y/////////////// 

/// 

/" 

IW 


T^ 


<- 


if; 


-i— 


Fig.  277 


directions  at  and  c„  respectively,  we  then  have  for  our  three  equations  of 
resolution 

HFt  =  Mlt,        2F„  =  Man,        2^3  =  o, 

where  H^Ft,  SF^,  and  IIF3  stand  for  the  algebraic  sums  of  the  components  of 
all  the  external  forces  acting  on  the  body,  along  the  three  lines  of  resolution 
just  named. 

Examples.  —  i.  Fig.  277  represents  a  flat  car  on  which  there  is  a  body  A 
weighing  4000  pounds.     Suppose  that  the  car  is  on  a  curve  with  no  elevation 

,  of  outer  rail;   that  the  speed  of  the  car  is 

increasing  at  2  miles  per  hour  per  second. 
Required  the  reaction  of  the  car  on  A  at 
the  instant  when  the  velocity  is  40  miles 
per  hour,  and  where  the  radius  of  the 
curve  is  1000  feet.  There  are  two  forces 
acting  on  A,  its  weight  and  the  reaction 
of  the  car.  For  simplicity  we  imagine  the 
reaction  resolved  into  two  components,  normal  pressure  N  (vertical)  and 
friction  F  (horizontal);  next  we  imagine  F  resolved  into  two  components, 
along  the  tangent  and  the  radius  of  the  path  of  the  center  of  mass  of  A ,  and 
we  call  them  Fi  and  Fo  respectively  (see  Fig.  277),  where  A  is  shown  in  plan 
and  elevation.     Therefore 

2F,  =  Fi  =  Mat,    SF„  =  F2  =  Mon,    and    EF3  =  N  -  4000  =  o. 

Now  at  =  2  miles  per  hour  per  second  =  2.93  feet  per  second  per  second. 
The  velocity  is  40  miles  per  hour  or  58. 7  feet  per  second,  and  therefore  a„  = 
58.7^  -^-  1000  =  3.44  feet  per  second  per  second.  The  force-acceleration 
equations  become  Fi  =  (4000  -f-  32.2)  2.93  =  364  pounds,  F2  =  (4000  H- 
32.2)  3.44  =  427  poimds,  and  N  =  4000  pounds.     The  reaction  sought  equals 

V'(4ooo2  +  364-  +  427-)  =  4039  pounds. 

We  have  assumed  that  A  and  the  floor  of  the  car  are  rough  so  as  to  furnish  a 
frictional  force  large  enough  to  hold  A  in  place  on  the 
car;  the  necessary  holding  force  =  V(364^  +  427^)  = 
561  pounds. 

2.  A  circular  cylinder  C  (Fig.  278)  is  laid  in  a  box 
which  is  mounted  on  a  board  as  shown,  and  the  whole 
thing  is  then  rotated  about  a  vertical  axis  AB.  The 
weight  of  the  cylinder  is  30  pounds,  AC  =  2  feet,  and 
the  rate  of  rotation  (constant)  =  60  revolutions  per 
minute.     The  pressures  of  the  box  on  the  cyHnder 

are  required.  There  are  three  forces  acting  on  the  cylinder,  —  its  weight, 
the  pressure  Pi  of  the  bottom  of  the  box,  and  a  pressure  P2  exerted  by 
one  of  the  ends  of  the  box.  We  assume  that  the  cylinder  rests  against  the 
lower  end;    the  complete  solution  will  determine  whether  the  assumption  is 


Fig.  278 


Art.  34 


i6i 


correct.  Because  the  rate  of  rotation  is  constant  there  are  no  pressures  on  the 
cylinder  in  the  direction  of  motion  (perpendicular  to  paper).  The  velocity 
of  the  mass-center  =27r2X6o=754  feet  per  minute  =  12.5  feet  per  second; 
hence  fl  =  78  feet  per  second  per  second,  directed  toward  the  axis  of  rotation. 
Now  11F„  =  Pi  sin  30°  —  P2  cos  30°  =  (30  -^-  32.2)  78,  and  'LF3  =  Pi  cos  30° 
+  P2  sin  30°  —  30  =  o.  Solving  them  simultaneously  for  Pi  and  P2  we  get 
Pi  =  62.3  and  P2  =  —  48.0  pounds.  The  negative  sign  means  that  we  made 
a  wrong  assumption  as  to  P2;  it  acts  downward  and  is  exerted  by  the  upper 
end  of  the  box. 

3.  A  simple  conical  pendulum  consists  of  a  "bob  "  suspended  from  a  fixed 
point  by  a  cord,  arranged  so  the  bob  and  cord  can  be  rotated  about  a  vertical 
through  the  fixed  point.  See  Fig.  279  which  represents  such  a  pendulum  by 
side  and  end  views;  /1 5  is  a  forked  vertical  shaft;  GG  are  guides  fastened  to 
the  shaft,  between  which  the  bob  may  swing.  When  the  shaft  is  rotated,  the 
cord  will  deflect  from  the  vertical.  We  now  determine  this  deflection  for  any 
constant  speed  of  rotation.  Let  /  =  length  of  cord,  from  point  of  suspension 
to  the  center  of  the  bob;  6  =  angle  of  deflection;  n  =  number  of  revolutions 
per  unit  time;  W  =  weight  of  bob;  T  —  tension  in  the  cord.  The  bob  is 
under  the  action  of  W,  T,  and  the  pressure  P  of  one  of  the  guides  possibly; 
hence 

2P„  =  T  sin  e  =  AF^n-,    SF3  =Tcosd-W=  o;    ^Ft  =  P  =  Mat. 

When  the  speed  is  constant  as  here  assumed,  the  deflection  is  constant,  and  the 
center  of  the  bob  describes  a  horizontal  circle  of  radius  /  sin  6.  The  velocity 
of  the  center  =  2  tt  /  sin  0  •  n ;  hence  a„  =  4  ir'^nH  sin  6,  and  P  sin  0  =  (^V/g) 
4  ir'^nH  sin  6.  Solving  this  and  T  cos  6  =  W  simultaneously  for  d  we  get 
cos^  =  g  -^  {4Tr^nH).     Also  T  =  W  4-ir^nH  -^  g;  and  since  at  =  o,  P  =  o. 


Fig.  280 


"  on  a  curve  "  in 


Elevation  of  Outer  Rail  on  Curves.  —  Fig.  280  represents  a  car 
a  railway  track.  We  discuss  certain  features  of  the  pressures  of  the  car  upon 
the  track  as  the  car  runs  around  the  curve.  Imagine  the  rail  pressure  on  each 
wheel  resolved  into  three  components,  —  one  parallel  to  the  ties  (so-called 
flange  pressure),  one  perpendicular  to  the  track,  and  one  parallel  to  the  rails. 
Unless  the  curve  is  quite  sharp,  the  forces  of  each  of  these  three  sets  of  com- 
ponents are  parallel.    We  will  suppose  them  parallel,  and  call  the  resultants 


l62  Chap,  vin 

of  the  three  sets  Ri,  R2,  and  R3  respectively.  Besides  these  three  resultants 
there  are  acting  on  the  car  the  weight  W,  the  pull  Pi  of  the  car  ahead,  and  the 
pull  P2  of  the  car  behind.  Unless  the  curve  is  quite  sharp  Pi  and  P2  are  practi- 
cally parallel  to  the  tangent  to  the  curve  under  the  middle  of  the  car;  we  will 
assume  them  to  be  so.  Then  resolving  along  the  normal  (or  radius  of  the 
curve),  the  vertical,  and  the  tangent  to  the  curve,  we  get 

Ri  cos  <^  +  i?2  sin  0  =  (W  ^  g)  a„  =  (PF  -4-  g)  v^/r, 

where  v  =  velocity  of  the  car  and  r  =  radius  of  the  curve; 

-  Ri  sin  <p -\- R2  cos  (f)  -  W  =  o;    and    Pi  -  Po  -  i?3  =  (TT  -J-  g)  at. 

Solving  the  first  and  second  sunultaneously  for  Ri  and  R2  we  get 

72,  =  T^  f  —  cos  <i  —  sin  d) ) ,     and     P2  =  W  ( —  sin  0  +  cos  ^  I  • 
\gr       ^  I  \gr  I 

It  is  obvious  from  the  expression  for  Ri  that  the  resultant  flange  pressure  may 
be  equal  to  zero  for  certain  values  of  v,  r,  and  0.  It  will  be  zero  if  {v-  cos  4>) 
-^  gr  =  sin  <i>,  or  ' 

tan  0  =  v^/gr* 

*  This  formula,  or  some  modification  of  it,  is  used  to  determine  the  proper  elevation  of  the 
outer  rail  on  railroad  curves,  except  as  noted  below.  The  following  is  a  practical  rule  deduced 
from  the  formula:  "The  correct  superelevation  for  any  curve  is  equal  to  the  middle  ordinate 
of  a  chord  [of  the  curve]  whose  length  in  feet  is  1.6  times  the  speed  of  the  train  in  miles  per 
hour."  On  the  Pennsylvania  Railroad  the  rule  is  modified  as  follows:  "  No  speed  greater 
than  50  miles  per  hour  should  be  assumed  in  determining  the  superelevation  by  the  above 
method  even  though  higher  speed  may  be  made.  No  superelevation  exceeding  7  inches  is 
permissible  and  none  exceeding  6  inches  should  be  used  except  at  special  locations  on  passenger 
tracks."  The  formula  was  deduced  on  the  basis  that  resultant  flange  pressure  should  =  zero. 
The  same  formula  is  arrived  at  by  making  ties  of  the  track  perpendicular  to  the  resultant 
pressure  between  the  floor  of  the  car  and  any  object  resting  upon  it,  or  perpendicular  to  a 
plumb  Une  suspended  in  the  car. 


-    P 


CHAPTER   IX 

TRANSLATION    AND    ROTATION 
35.   Translation 

A  translation  is  such  a  motion  of  a  rigid  body  that  each  straight  Hne  of  the 
body  remains  fixed  in  direction;  there  is  no  turning  about  of  any  line  of  the 
body.  The  coupling  or  side  rods  of  a  locomotive  (connecting  the  driving 
wheels  on  either  side)  have  a  translatory  motion  when  the  locomotive  is 
running  on  a  straight  track.  It  should  be  noticed  that  our  definition 
does  not  require  rectilinear  motion  of  each  point  of  the  moving  body.  But 
rectilinear  translations  are  most  common,  and  such  translations  have  been 
quite  fully  discussed  in  Art.  31. 

The  motions  of  all  points  of  a  body  in  translation  are  alike.  For,  let  A  and 
B  be  any  two  points  of  the  body,  and  A'  and  B'  be  the  positions  of  those  points 
in  space  at  a  certain  instant  and  A"  and  B"  their  positions  at  a  later  instant. 
By  definition  of  translation  the  Hnes  A'B'  and  A"B"  are  parallel;  and  since 
the  lines  are  equal  in  length  the  figure  A'B'  B"A"  is  a  parallelogram,  and 
A' A"  and  B'B"  (the  displacements  of  A  and  B  respectively)  are  equal  and 
parallel.  Since  the  displacements  of  all  points  of  the  moving  body  for  any 
interval,  long  or  short,  are  equal  and  parallel,  the  velocities  of  all  points  at  any 
instant  are  alike,  and  hence  also  the  accelerations.  By  displacement,  velocity, 
and  acceleration  of  a  body  having  a  motion  of  translation  is  meant  the  dis- 
placement, velocity,  and  acceleration  respectively  of  any  one  of  its  points. 

The  general  principle  of  Art.  34,  relating  to  the  motion  of  the  mass-center  of  a 
body  moving  in  any  way,  when  applied  to  a  translation,  takes  this  form:  the 
algebraic  sum  of  the  components  —  along  any  line  —  of  the  external  forces 
acting  on  the  body  equals  the  product  of  the  mass  of  the  body  and  the  com- 
ponent of  the  acceleration  of  the  body  along  that  line.  This  gives  three  in- 
dependent "equations  of  motion,"  namely, 

2/^x  =  Ma.,        ^Fy=  May,        2F.  =  Ma„ 

where  x,  y,  and  z  denote  three  noncoplanar  lines  of  resolution. 

The  resultant  of  all  the  external  forces  acting  on  a  body  having  a  motion  of  trans- 
lation is  a  single  force;  its  line  of  action  passes  through  the  mass-center,  the  force 
is  directed  like  the  acceleration  of  the  body,  and  its  magnitude  equals  the  product 
of  the  mass  of  the  body  and  the  acceleration.*  Assuming  that  the  resultant  is  a 
single  force,  most  students  will  accede  to  the  second  statement  in  the  foregoing 

*  The  student  is  reminded  that  the  resultant  of  a  system  of  forces  is  a  force,  a  couple,  or  a 
pair  of  noncoplanar  forces  (see  Chapter  I). 

163 


164 


Chap,  ix 


proposition,  on  the  basis  of  their  experience;  for,  they  will  say,  if  the  resultant 
did  not  pass  through  the  mass-center,  the  body  would  turn  and  not  have  a 
translatory  notion.  But  it  can  be  demonstrated  as  follows:  Let  Fig.  281 
represent  the  body  and  points  i,  2,  3,  etc.,  its  constituent  particles;  the 
external  forces  acting  on  the  body  are  not  shown.  Suppose  that  the  accelera- 
tion is  directed,  say,  toward  the  right,  and  let  a  =  the  magnitude  of  that 
acceleration,  and  m\,  W2,  m^,  etc.  =  the  masses  of  the  particles  respectively. 
Then  the  resultants  of  all  the  forces  acting  on  the  several  particles  equal  re- 
spectively mia,  in^a,  m^a,  etc.,  all  directed  like  the  acceleration,  as  represented 
in  the  figure.  Now  this  system  of  imaginary  forces  (resultants)  is  equivalent 
to  all  the  real  forces,  external  and  internal,  acting  on  the  system  of  particles; 
and  the  resultant  of  the  imaginary  system  and  that  of  the  real  system  are 
identical  in  magnitude,  line  of  action,  and  sense.  But  the  internal  forces  occur 
in  pairs  of  equal,  colinear,  and  opposite  forces  (Art.  34),  and  so  constitute  a 
balanced  system  and  contribute  nothing  to  the  resultant  of  the  real  system. 
Hence,  the  resultant  of  the  external  system  and  that  of  the  imaginary  system 
are  identical.  We  proceed  now  to  ascertain  the  resultant  from  the  latter 
system. 

A 


Fig.  23i 


Fig.  282 


The  imaginary  system  (I)  consists  of  parallel  forces  proportional  to  the 
masses  of  the  particles,  and  the  lines  of  action  of  the  forces  pass  through  the 
particles  respectively.  The  system  of  earth-pulls  (gravity,  G)  likewise  con- 
sists of  parallel  forces  proportional  to  the  masses  of  the  particles,  and  the  lines 
of  action  of  these  pulls  pass  through  the  particles  respectively.  Hence  systems 
T  and  G  are  very  similar;  and  if  we  imagine  the  body  turned  so  that  the  line 
AB  (parallel  to  a)  in  Fig.  281  is  vertical  (Fig.  282)  then  systems  /  and  G  are  still 
more  alike.  The  difference  is  in  the  magnitudes  of  corresponding  forces;  the 
forces  of  /  are  respectively  proportional  to  the  forces  of  G.  It  follows  that  the 
line  of  action  of  the  resultants  of  systems  /  and  G  coincide  (in  the  body) ;  but  the 
resultant  of  system  G  passes  through  the  mass-center  of  the  body;  and  hence 
the  resultant  of  system  /  (and  the  resultant  of  the  external  system)  also  passes 
through  the  mass-center.  From  Fig.  281  it  is  obvious  that  the  resultant  of  the 
external  system  is  a  single  force  directed  like  the  acceleration,  and  equals 

Wifl  -\-  m2(i  +  •  •  •   =  o^m  =  Ala. 

The  algebraic  sum  of  the  moments,  or  torque,  of  all  the  external  forces  about  any 
line  through  the  mass-center  equals  zero,  for  the  resultant  of  those  forces  has  no 


Art.  35 


165 


ZOOOIbs. 


moment  about  such  line.  This  principle  gives  three  independent  m.oment 
equations: 

r.  =  o,      r,  =  o,      r.  =  o,  (i) 

where  T^,  Ty,  and  Tz  denote  the  moment-sums  for  three  noncoplanar  lines 
through  the  mass-center.  Or  we  may  take  moments  about  any  three  lines 
and  equate  the  torques  of  the  external  forces  about  the  lines  to  the  moments 
of  the  resultant  {Ma)  about  the  same  lines  respectively. 

Examples.  —  i.  A  (Fig.  283)  is  a  rectangular  prism  weighing  2000  pounds. 
The  car  is  being  started  at  4  feet  per  second  per  second.  Required  the  pressure 
of  the  car  on  the  bottom  of  the  prism.  There 
are  only  two  forces  acting  on  the  prism,  —  its 
own  weight  and  the  required  pressure  P.  See 
the  figure  where  P  is  shown  resolved  into  two 
components  (Pi  and  P2)  at  the  base  of  the  prism. 
The  (unknown)  distance  from  the  point  of  ap- 
plication of  P  to  the  center  of  the  base  is  de- 
noted by  X.  HFy  =  Pi  —  2000  =  May  —  o,  or 
Pi  =  2000;  HF^  =  P2  =  (2000/32.2)  4  =  248.  Hence  P  =  V(2ooo2 -|- 248^)  = 
2015  pounds,  and  the  inclination  of  P  to  the  vertical  =  tan~^  (248/2000)  = 
8°  25'.  To  determine  x  we  take  the  torque,  of  the  forces  acting  on  the  prism, 
about  the  horizontal  line  through  the  mass-center  and  perpendicular  to  the 
direction  of  motion  and  equate  to  zero.  Thus  248  X  2.5  —  2000  x  =  o,  or 
X  =  0.31  feet  =  3.72  inches.  (P2  =  24S  pounds  is  friction,  and  the  floor  and 
prism  must  be  rough  enough  to  develop  such  a  value,  to  prevent  the  slipping, 
here  assumed  not  to  occur.  Thus  the  coefficient  of  friction  must  be  not  less 
than  248  -^  2000  =  0.124  or  about  one-eighth.  If  the  coefiicient  were  less 
than  one-eighth,  the  friction  developed  under  the  prism,  say  200  pounds,  could 
not  give  the  prism  an  acceleration  of  4  feet  per  second  per  second,  only  3.22. 
Hence  the  prism  would  eventually  be  left  behind.  The  prism  is  not  "thrown 
off  by  the  force  of  inertia  "  in  such  a  case,  as  some  would  describe  the  phe- 
nomenon, but  the  car  slips  out  from  under  the  prism.) 

2.   C  and  C  (Fig.  284)  are  two  parallel  cranks,  their  shafts  being  connected 
mechanically  so  that  they  rotate  together  with  equal  speeds  and  in  the  same 

direction.  P  is  a  bar  pinned  to  the  cranks.  We 
discuss  the  forces  acting  on  B  when  the  mechanism 
is  in  motion.  There  are  three  such  forces ;  the  weight 
of  B  and  the  pressures  of  the  pins  on  B.  We  will 
neglect  the  weight,  or  assume  that  the  plane  of  the 
cranks  is  horizontal  so  that  the  bar  lies  upon  the 
cranks  and  the  supporting  forces  balance  the  weight. 
If  the  bar  is  uniform  then  it  seems  reasonable  to 
assume  that  the  pin  pressures  Q  are  parallel;  if  so  they  must  be  equal  since 
the  algebraic  sum  of  their  moments  about  the  mass-center  of  B  equals  zero. 


-/^ 


^ 


^^^ 


/^ 


Fig.  284 


1 66 


Chap,  ix 


Moreover,  the  resultant  of  the  pin  pressures  =  <3  +  Q  =  Ma,  where  M  =  mass 
of  the  rod  and  a  =  its  acceleration,  and  the  pressures  act  in  the  direction  of  a. 
The  acceleration  of  the  bar  is  the  same  as  that  of  the  center  of  either  pin  P. 
If  the  cranks  be  made  to  turn  uniformly,  then  the  acceleration  is  in  the  direc- 
tion PO  and  it  equals  V"/r  (Art.  32),  where  v  =  velocity  of  P  and  r  —  PO; 
hence  2  Q  =  Mv^/r  =  (W/g)  {fir),  or(^  =  \  Wv^gr. 

3.  Imagine  a  locomotive  raised  up  off  its  track,  and  that  steam  is  "turned 
on"  so  that  the  drivers  are  made  to  rotate  at  constant  speed.  If  the  connect- 
ing rod  on  one  side  be  detached  —  the  drivers  being  driven  from  the  other 
side  —  then  the  side  rod  on  the  first  side  would  be  under  the  action  of  pin 
pressures  just  like  those  discussed  in  the  preceding  example.  Each  pressure 
equals  |  Mv^/r,  directed  along  its  crank  radius  and  toward  the  crank  shaft. 
(The  weight  of  the  rod  induces  pressures  equal  to  |  PF  upwards.) 

When  the  locomotive  is  running  on  its  track,  then  there  is  superimposed 
upon  the  motion  of  the  side  rod  just  discussed  the  forward  (or  backward) 
motion  of  the  locomotive  as  a  whole.  The  velocity  of  the  side  rod  equals  the 
vector  sum  of  v  and  the  velocity  of  the  locomotive ;  and  the  acceleration  of  the 
rod  equals  the  vector  sum  of  the  acceleration  V'/r  and  that  of  the  locomotive. 
Now  when  the  velocity  of  the  locomotive  is  constant  its  acceleration  is  zero, 
and  the  acceleration  of  the  side  rod  is  still  V"/r  and  parallel  to  the  cranks  and 
directed  as  explained  in  example  2,  Hence,  even  when  the  locomotive  is 
running  on  a  track,  the  pin  pressures  on  the  (lone)  side  rod  are  as  when  the 
locomotive  is  "jacked  up  "  and  running.  Let  V  =  speed  of  locomotive,  R  = 
radius  of  driving  wheels;  then  v  =  Vr/R,  and  the  pin  pressures  =  ^  (W/g)  r 
V^/R^  (weight  of  rod  neglected).  For  example,  let  W  =275  pounds,  r  —  i 
foot,  R  =  2.75  feet,  and  F  =  60  miles  per  hour  =  88  feet  per  second;  then 
the  pin  pressures  =  ^  (275/32.2)  X  i  X  (88  -^  2.75)^  =  4425  pounds. 

Locomotive  Side  Rod.  —  We  give  here  another  solution  of  the  side  rod  prob- 
lem (see  preceding  examples).     In  Fig.  285  each  pin  pressure  on  the  rod  is 

represented  by  two  components,  hori- 
zontal and  vertical.  The  vertical  com- 
ponents are  equal  since  the  sum  of  the 
moments  of  all  the  forces  acting  on  the  rod 
(pressures  and  weight)  about  the  center 
of  gravity  (at  mid-length  of  the  rod) 
equals  zero;  hence  both  vertical  com- 
ponents are  denoted  by  the  same  letter 
The  horizontal  components  are  A^i  and  A2.  Let  a  =  the  total,  or  abso- 
lute, acceleration  of  any  and  every  point  of  the  rod  when  the  cranks  make  any 
angle  6  with  the  downward  vertical,  and  cz^  and  ay  =  the  horizontal  and  verti- 
cal components  of  a.     Then 

■X":  —  X2  =  Max,     and 


Fig.  285 


Y. 


2Y-W  =  May,    or     Y  =  ^W  +  May). 


Presently  we  show  how  to  find  ax  and  ay  for  any  position  of  the  cranks.     Then 


Art.  35 


167 


from  the  above  we  can  determine  A'l  -  X2  and  F.  The  values  of  Xx  and  Xg 
depend  upon  the  load  or  pull  on  the  locomotivCj  and  how  it  is  distributed 
among  the  driving  wheels.  But  Y  does  not  depend  on  the  pull,  only  on  W 
and  Qy. 

We  now  discuss  the  motion  of  one  of  the  crank  pins  with  the  view  of  obtain- 
ing formulas  for  a^  and  Oy.  Let  V  =  the  velocity  of  the  locomotive,  A  =  its 
acceleration,  R  =  radius  of  the  driving  wheels,  and 
r  =  length  of  the  cranks  {CP,  Fig.  286).  It  will 
be  convenient  to  refer  the  motion  of  the  crank-pin 
P  to  the  coordinate  axes  shown;  OY  is  the  position 
occupied  by  the  crank  when  P  was  in  its  lowest 
position.  Let  5  be  the  distance  of  C  from  OY,  and 
X  and  y  the  coordinates  of  P.     Then 


and 


s  =  Rd, 

X  =  s  —  r  sind, 

y  =  R  —  r  cos  6 


Fig.  286 


Now  a^  =  d'^x/dl'^  and  Qy  =  d-y/dt"^,  and  for  use  below  V  =  ds/dt  =  R  dd/dt,  or 
dd/dt  =  V/R.    Thus 


dx      ds  ^   dd      ,,  V 

-  =  --rcose^-=V-rcose.-^V^i 


-  ^  cos  dj ; 


dVf         r        ^  ,  r.r    .    ^    dd 

'^^^V-r'''V-^^R''''^'dt 


-■<^--u 


R^ 


^cosdj-\-—rsme; 


dV  r    .  r  dd  r  V^ 


Thus  it  is  seen  that  a^  and  Oy  depend  on  the  velocity  and  acceleration  of  the 
locomotive.  The  largest  values  of  c^  and  a„  obtain  at  high  speed,  and  then  the 
A  terms  (in  the  expressions  for  a^  and  ay)  are  small  and  negligible  compared  to 
the  V  terms.  So  when  we  neglect  these  terms  or  when  the  acceleration  of 
the  locomotive  is  zero,  then 

a^  =  (V/Py  r  sin  6,     and     ay  =  (V/Py  r  cos  6. 

When  the  rod  is  in  its  lowest  position,  6^0,0^  =  0,  a-y  =  {V/RYr,  Xi  =  X2, 
and  F  =  I  W  -^  ^  (W/g)  (F/7?)V;  the  forces  F-  act  upward  on  the  rod. 
When  6  =  90°,  a^  =  (V/R)h,  a^  =  o;  the  resultant  of  the  two  forces  X  acts 
toward  the  right  and  equals  (W/g)  (V/R)h,  and  F  =  |  W.  When  the  rod  is 
in  its  highest  position  d  =  180°,  a^  =  o,  ay  =  -  {V/R)h;  Xi  =  X2,  and  F  =  | 
W  -^  (W/g)  {V/R)h;  for  high  speeds  F  acts  down  on  the  rod.  When 
6  =  270°,  flj  =  —(V/R)h,  Qy  =  o;  the  resultant  of  Xi  and  Xi  acts  toward  the 
left  and  equals  {W/g)  {V/R)h,  and  F  =  |  W. 


s 


r 


1 68  Chap,  ix 

36.   Moment  of  Inertia  and  Radius  of  Gyration 

§1.  General  Principles,  Etc.  —  Perhaps  every  student  has  observed 
that  the  effort  required  to  start  a  body  to  rotating  about  a  fixed  axis  seems  to 
depend  not  only  on  the  mass  of  the  body  but  also  on  the  remoteness  of  the 
material  of  the  body  from  the  axis  of  rotation.     Fig.  287  represents  a  simple 

apparatus  by  means  of  which  one  can  roughly  "sense  " 

-^   this  fact.     It  consists  of  a  vertical  shaft  5  to  which  a 

grooved  pulley  P  and  cross  arm  A  are  fastened  rigidly, 

and  a  heavy  body  B  which  can  be  clamped  on  the  cross 

arm.     The  pull  or  turning  effort  may  be  applied  by 

means  of  a  cord  wrapped  about  the  pulley.     It  is  shown 

Fig.  287  .^  ^^^  following  article  that  this  "rotational  inertia  " 

of  a  body  is  proportional  to  the  "moment  of  inertia  "  of  the  body  about  the 

axis,  and  this  article  is  devoted  to  a  discussion  of  moment  of  inertia,  as  a 

preparation  for  the  following  article.* 

The  moment  of  inertia  of  a  body  with  respect  to  a  line  is  the  sum  of  the  prod- 
ucts obtained  by  multiplying  the  mass  of  each  particle  of  the  body  by  the 
square  of  its  distance  from  the  line.  Or,  if  /  =  moment  of  inertia,  mi,  mi,  m^, 
etc.  =  the  masses  of  the  particles,  and  n,  r^,  r^,  etc.,  their  distances  respectively 
from  the  line  or  axis,  then 

/  =  miTx^  +  nhr2^  +  •  •  •   =  Swr^; 
or  if  the  body  is  continuous,  then 

/=   jdM-r^,  (i) 

where  dM  denotes  the  mass  of  any  elementary  portion  and  r  its  distance  from 
the  line  about  which  moment  of  inertia  is  taken.  The  elementary  portion 
must  be  chosen  so  that  each  point  of  it  is  equally  distant  from  the  line, 
else  there  is  doubt  as  to  what  distance  to  take  for  r. 

It  is  plain  from  the  foregoing  formulas  that  a  unit  of  moment  of  inertia  de- 
pends upon  the  units  of  mass  and  distance  used.  There  is  no  single-word  name 
for  any  unit  of  moment  of  inertia.  Each  unit  is  described  by  stating  the  units 
of  mass  and  distance  involved  in  it,  and  in  accordance  with  the  "make-up  " 

*  Euler  (1707-83)  introduced  the  term  "moment  of  inertia,"  and  he  explained  its  appro- 
priateness (in  his  "Theoria  Motus  Corporum  SoHdorum,"  p.  167)  somewhat  as  follows:  The 
choice  of  the  name,  moment  of  inertia  (Ger.  tragheits-moment),  is  based  on  analogies  in  the 
equations  of  motion  for  translations  and  rotations.  In  a  translation  the  acceleration  is 
proportional  directly  to  the  "accelerating  force"  and  inversely  to  the  mass,  or  "inertia," 
of  the  moving  body;  and  in  a  rotation  the  angular  acceleration  is  proportional  directly  to 
the  moment  of  the  accelerating  force  and  inversely  to  a  quantity,  Xmr~,  depending  on  the 
mass  or  inertia.  This  quantity,  to  complete  a  similarity,  we  may  call  "moment  of  inertia." 
Then  we  have  for  translations  and  rotations  respectively, 

linear  acceleration      =  (force) /(inertia  or  mass);  and 

angular  acceleration  =  (moment  of  force)/(moment  of  inertia). 


Art.  36  169 

of  the  unit.     Thus,  when  the  pound  and  the  foot  are  used  as  units  of  mass  and 

length  respectively,  then  the  unit  of  moment  of  inertia  is  called  a  pound-foot 

square;    when  the  slug  (about  32.2  pounds)  and  the  foot  are  used,  then  the 

unit  moment  of  inertia  is  called  slug-foot  square.* 

The  moment  of  inertia  of  any  right  prism  —  cross  section  of  any  form  — 

with  respect  to  any  line  parallel  to  the  axis  of  the  prism  can  be  computed  in  a 

special  way,  preferred  by  some.     Thus  if  we  take  as  elementary  portion  a 

filament  of  the  prism  parallel  to  the  axis,  then  dM  =  {adA)  8  where  a  =  the 

altitude  of  the  prism,  dA  —  the  cross  section  of  the  filament,  and  8  =  density; 

and  p 

1=  a8  I  dAr\  (2) 

This  integral  (extending  over  the  area  of  the  cross  section)  is  called  the  moment 
of  inertia  of  the  cross  section  about  the  line  specified  (see  appendix  B). 

Since  a  moment  of  inertia  is  one  dimension  in  mass  and  two  in  length,  it  can 
be  expressed  as  the  product  of  a  mass  and  a  length  squared;  it  is  sometimes 
convenient  to  so  express  it.  The  radius  of  gyration  of  a  body  with  respect  to 
a  line  is  such  a  length  whose  square  multiplied  by  the  mass  of  the  body  equals 
the  momxent  of  inertia  of  the  body  with  respect  to  that  line.  That  is,  if  k  and  / 
denote  the  radius  of  gyration  and  moment  of  inertia  of  the  body  with  respect 
to  any  axis  and  M  =  its  mass,  then 

¥M  =  I    or     k=  y/JjM.  (3) 

The  radius  of  gyration  may  be  viewed  as  follows :  If  we  imagine  all  the  material 
of  a  body  concentrated  into  a  point  so  located  that  the  moment  of  inertia  of 
the  material  point  about  the  line  in  question  equals  the  moment  of  inertia  of 
the  body  about  that  line,  then  the  distance  between  the  line  and  the  point 
equals  the  radius  of  gyration  of  the  body  about  that  line.  The  material  point 
is  sometimes  called  the  center  of  gyration  of  the  body  for  the  particular  line. 

To  furnish  still  another  view  of  radius  of  gyration  we  call  attention  to  the 
fact  that  the  square  of  the  radius  of  gyration  of  a  homogeneous  body  with 
respect  to  any  line  is  the  mean  of  the  squares  of  the  distances  of  all  the  equal 
elementary  parts  of  the  body  from  that  line.  For  let  r^,  r^,  etc.,  be  the  dis- 
tances from  the  elements,  dM,  to  the  axis,  and  let  n  denote  their  number  (in- 
finite).    Then  the  mean  of  the  squares  is 

W  +  rs^  -f  •••)/«  =  {n^dM  +  r^^dM  +  -  •  •  )/ndM  =  I/M  =  k\ 

Obviously  the  radius  of  gyration  of  a  body  with  respect  to  a  line  is  intermediate 
between  the  distances  from  the  line  to  the  nearest  and  most  remote  particle  of 
the  body.  This  fact  will  assist  in  estimating  the  radius  of  gyration  of  a  body. 
Examples.  —  i.  Required  to  show  that  the  moment  of  inertia  of  a  slender 
rod  about  a  line  through  the  center  and  inclined  at  an  angle  with  the  rod  is 
-x^MP  sin^  a,  where  M  =  mass,  I  =  length,  and  a  =  angle  between  the  line  and 
the  axis  of  the  rod.     Let  a  —  the  cross  section  of  the  rod,  5  =  the  density,  and 

*  For  dimensions  of  a  unit  of  moment  of  inertia,  see  Appendix  A. 


lyo 


Chap,  ix 


X  =  the  distance  of  any  elementary  portion  from  the  middle  of  the  rod  AB 
(Fig.  288).  Then  dM  =  5  {a  dx),  and  the  distance  of  the  element  from  CD  = 
X  sin  a.     Hence 

I  =    I         8adx'  x^  sm^  a  =  8a  sm^  a   —  =  ■ ' 

J-U  ISA-u  3        4 

and  this  reduces  to  yV  MP  sin^  a,  since  8al  =  M. 


Fig.  288 


Fig.  289 


2.  Required  to  show  that  the  moment  of  inertia  of  a  right  parallelopiped 
about  a  central  axis  parallel  to  an  edge  equals  yV  M  (a-  +  b-)  where  M  =  mass 
of  the  parallelopiped  and  a  and  b  =  the  lengths  of  the  edges  which  are  perpen- 
dicular to  that  axis.  See  Fig.  289  where  the  s  axis  is  the  one  to  which  this 
moment  of  inertia  corresponds.  We  take  for  elementary  portion  a  volume 
dxdydz;  its  mass  =  8  (dxdydz),  and  the  square  of  its  distance  from  the 
z  axis  =  x^  +3'^.     Hence 

I  I    (x2  +  v2)  dx  dy  dz=—  {a?b  +  a¥)  =  etc. 

•a/2    J -b/2    Jo  12 

3,  Required  to  show  that  the  moment  of  inertia  of  a  right  circular  cylinder 
with  respect  to  its  axis  is  |  Mr"^,  where  M  =  its  mass  and  r  =  the  radius  of  its 
base.  We  use  the  special  method  for  prisms  (see  equation  2)  and  choose  polar 
coordinates  (see  Fig.  290) ;  then  dA  =  pdddp  and  dM  =  8  {ap  dd  dp) ;  hence 

I=a8  fdAp'  =  a8   r     P^^  dp , '"      ''^'^' "" 


'de 


=  etc. 


Fig.  290 


Fig.  291 


4.  Required  to  show  that  the  moment  of  inertia  of  a  sphere  about  a  diameter 
is  I  Mr^  where  M  =  its  mass  and  r  =  its  radius.  We  might  begin  with  equation 
(i)  but  we  will  use  a  special  method,  making  use  of  the  result  found  in  exam- 
ple 3.  We  conceive  the  sphere  made  of  laminas  perpendicular  to  the  diameter 
in  question;  determine  the  moment  of  inertia  of  each  lamina;  and  then  add  the 
moments  of  inertia  of  the  laminas.     Let  XX'  (Fig.  291)  be  the  diameter  in 


Art.  36 


171 


question,  and  PQ  a  section  of  one  of  the  laminas;  then  the  mass  of  the  lamina 
is  b  (iry'^dx).  According  to  example  3  the  moment  of  inertia  of  this  lamina 
(cylinder)  about  its  axis  (XX')  is  ^  5  (iry^  dx)y^.  Hence  the  moment  of  inertia 
of  the  sphere  is 

J       A  5  (tt/  dx)  =  ^  ttS  J_     (r2  -  ^2)2  dx  =  j%  dw  r^  =  etc. 

§  2.  Parallel  Axis  Theorem;  Reduction  or  Transformation  Formu- 
las.—  There  is  a  simple  relation  between  the  moments  of  inertia  (and  the 
radii  of  gyration)  of  a  body  with  respect  to  parallel  lines  one  of  which  passes 
through  the  mass-center  of  the  body.  By  means  of  this  relation  we  can 
simplify  many  calculations  of  moment  of  inertia,  and  avoid  integrations  (see 
examples  following);  it  may  be  stated  as  follows: 

The  moment  of  inertia  of  a  body  with  respect  to  any  line  equals  its  moment  of 
inertia  with  respect  to  a  parallel  line  passing  through  the  mass-center  plus  the 
product  of  the  mass  of  the  body  and  the  square  of  the  distance  between  the  lines. 
Or,  if  /  =  the  first  moment  of  inertia,  7  =  the  second  (for  the  line  through  the 
mass-center),  M  =  mass,  and  d  =  the  distance  between  the  parallel  lines, 

I  =  l+Md\  (4) 

Proof.  —  Let  0  (Fig.  292)  be  the  mass-center,  and  P  any  other  point  of  the 
body  (not  shown),  LL  the  line  about  which  the  mo- 
ment of  inertia  is  /,  and  OZ  a  parallel  line  (through 
the  mass-center)  about  which  the  moment  of  inertia 
is  /.     Distance   between   these  parallel   lines   is  d. 
For  convenience  we  take  x  and  y  axes  through  O, 
the  former  in  the  plane  of  the  two  parallel  lines  and 
the  latter  perpendicular  to  that  plane.     Let  x,  y,  and 
z  =  the  coordinates  of  P.     The  square  of  the  dis- 
tance of  P  from  the  z  axis  equals  x'^  +  y'^,  hence  7=   I  dM  (x"^ -\- y^) .    The 
square  of  the  distance  of  P  from  the  line  LL  equals  (d  —  x)^  +  y-,  hence 
/  =  J[{d  -  xy+y']dM  =  Jix^  +  y')dM  +  d'JdM  -2dJxdM. 

Now  the  first  of  the  last  three  integrals  =  /,  and  the  second  one  =  Md"^.  If 
now  we  show  that  the  third  =  o,  then  formula  (4)  is  proved.  The  third 
integral  is  proportional  to  the  moment  of  the  body  with  respect  to  the  yz 
plane;  but  this  plane  contains  the  mass-center,  and  hence  the  moment  equals 
zero  (Arts.  21  and  23).     Thus,  if  Tf  =  weight  of  the  body, 


Fig.  292 


JxdM=   CxdW/g  =  (i/g)Wx. 


If  we  divide  both  sides  of  equation  (4)  by  M,  we  get  I/M  =  I/M  -f  d^,  or 

k^  =  k^-\-  d^  (5) 


172  Chap.  IX 

that  is,  the  square  of  the  radius  of  gyration  of  a  body  with  respect  to  any  line 
equals  the  square  of  Us  radius  of  gyration  with  respect  to  a  parallel  line  passing 
through  the  mass-center  plus  the  square  of  the  distance  between  the  two  lines. 
According  to  (5)  k  is  always  greater  than  d;  that  is,  the  radius  of  gyration  of 
a  body  with  respect  to  a  line  is  always  greater  than  the  distance  from  the  line 
to  the  center  of  gravity  of  the  body.  But,  if  the  dimensions  of  the  cross 
sections  of  the  body  perpendicular  to  the  line  in  question  are  small  com- 
pared to  d,  then  k/d  is  small  compared  to  i ,  and  k  equals  d  approximately  (see 
example  2).  In  such  a  case  the  moment  of  inertia  is  approximately  equal 
to  Md\ 

Examples.  —  i.  Required  the  moment  of  inertia  of  a  prism  of  cast  iron 
(weighing  450  pounds  per  cubic  foot)  6  inches  X  9  inches  X  3  feet  with  respect 
to  one  of  the  long  edges.     The  block  weighs  507  pounds.     According  to  example 

2,  §  I,  the  moment  of  inertia  of  the  block  with  respect  to  the  line  through  the 
mass-center  parallel  to  the  long  edge  is  507  (6^  +  9-)  -^-  12  =  4940  pounds- 
inches^.  The  square  of  the  distance  from  a  long  edge  to  the  mass-center  = 
29.25  inches^;  hence  the  moment  of  inertia  desired  =  4940  -f  507  X  29.25 
=  19,760  pound-inches^  =  4.27  slug-feet-. 

2.  Required  the  radius  of  gyration  of  a  round  steel  rod  i  inch  in  diameter 
with  respect  to  a  line  1 2  inches  from  the  axis  of  the  rod.     According  to  example 

3,  §  I,  the  square  of  the  radius  of  gyration  of  the  rod  with  respect  to  its  axis  is 
5  0.5^  =  0.125  inches^.  According  to  equation  (5)  the  radius  of  gyration  desired 
=  V  (0.125  +  144)  =  12.01,  nearly  the  same  as  the  distance  from  the  line  of 
reference  to  the  mass-center  of  the  rod. 

3.  It  is  required  to  show  that  the  moment  of  inertia  of  a  right  circular  cone 
with  respect  to  a  line  through  the  apex  and  parallel  to  the  base  =  j\  M  {r^  + 
4  c^)  where  Af  —  mass  of  the  cone,  r  =  radius  of  its  base,  and  a  =  its  altitude. 

We  conceive  the  cone  as  made  of  laminas  parallel  to  the  base, 
find  the  moment  of  inertia  of  each  lamina  with  respect  to  the 
specified  line,  and  then  add  all  the  moments.  For  con- 
venience we  take  the  axis  of  the  cone  as  the  ^'-coordinate 
axis,  and  the  line  for  which  the  moment  of  inertia  is  required 
as  the  X  axis  (Fig.  293).  The  moment  of  inertia  of  the  lamina 
Fig.  293  indicated  about  a  diameter  is  j  dM  •  x"^  where  dM  =  the  mass 

of  the  lamina  and  x  =  its  radius.  Hence  its  moment  of  inertia  about  the  x 
axis  =  J  dMx"^  +  dMy"^  (see  equation  3),  and  the  moment  of  inertia  of  the  en- 
tire cone  =    /  {\dMx^  +  dMy^),  the  limits  being  assigned  so  as  to  include  all 

laminas.  We  choose  to  integrate  with  respect  to  y,  and  so  must  express  dM 
and  X  in  terms  of  y.  From  similar  triangles  in  the  figure  x/y  =  r/a,  or 
X  =  ry/a;  obviously  dM  =  dirx^dy  =  8t  (r^y^/a^)  dy  where  5  =  density.     Hence 

C  Trr^hfdy   ,      C'^Trr^v^dv      irr^Sa   ,   irr^a^ 

I  =    I     ^^  -f    /     V-^  = 1 — -  =  etc. 

Jo        4  a*  Jo         a  20  s 


l«—  r  -> 

Y 

*-x^/ 

a      \ 

1     \ 

/P 

/  ^ 

/     1    X 

Aet.  36 


173 


Composite  Body.  —  By  this  term  is  meant  a  body  which  one  naturally  con- 
ceives as  consisting  of  finite  parts,  for  example,  a  flywheel  which  consists  of  a 
hub,  several  spokes,  and  a  rim.  The  moment  of  inertia  of  such  a  body  with 
respect  to  any  line  can  be  computed  by  adding  the  moments  of  inertia  of  all 
the  component  parts  with  respect  to  that  same  line.  The  radius  of  gyration 
of  a  composite  body  does  not  equal  the  sum  of  the  radii  of  gyration  of  the 
component  parts.  It  can  be  determined  from  equation  (3),  where  /  = 
moment  of  inertia  of  the  whole  body  and  M  =  its  mass. 

§  3.  Radius  of  Gyration  of  Some  Homogeneous  Bodies. — Let  k  = 
radius  of  gyration,  a  subscript  with  k  referring  to  the  axis  with  respect  to  which 
k  is  taken;  thus  kx  means  radius  of  gyration  with  respect  to  the  x  axis.  Also 
M  =  mass  and  8  =  density. 

Straight  Slender  Rod.  —  Let  /  =  its  length,  a  =  angle  between  the  rod  and 
the  axis.  Then  about  an  axis  through  the  mass-center  k^  =  tV  l^  sin^  a; 
about  an  axis  through  one  end  of  the  rod  k^  =  ^  P  sin^  a. 

Slender  Rod  Bent  into  a  Circular  Arc  (Fig.  294). — Let  r  =  radius  of  the 
arc,  then 

kx-  ^  I ''"  [i  —  (sin  a  cos  a)/a\,  and  ky"^  =  |  r^  [i  +  (sin  a  cos  a) /a]. 

The  divisor  a  must  be  expressed  in  radians   (i   degree  =  0.0175   radians). 
j^2  =  ^2  (^^j^g  2  axis  is  through  O  and  perpendicular  to  the  plane  of  the  arc). 


0-H^ 


P'iG.  295 


Fig.  296 


Fig.  297 


Right  Paralleloplped  (Fig.  295).— The  axis  OX  contains  the  mass-center, 
and  is  parallel  to  the  edge  c;  kx^  —  tV  (^^  +  ^")- 

Right  Circular  Cylinder  (Fig.  296).  — Both  axes  OX  and  OY  contain  the 
mass-center,  r  =  radius  and  a  =  altitude;  then 

kx'^hr^     V  =  tV  (3 '-'  +  «')• 

Hollow  Right  Circular  Cylinder  (Fig.  297).  — Let  R  =  outer  radius,  r  =  inner 
radius,  and  a  =  altitude;  then  ^^^.^ iY 

kx'  =  h(R'  +  r');    ky'^  =  HR'  +  r^~  +  U')-  I 


Right   Rectangular  Pyramid  (Fig.   298). — The   x   axis  ^ 
contains  the  mass-center  and  is  parallel  to  the  edge  a;    ^__^_—^,- 
M  =  I  ahhb.  Fig.  298 


iy4  Chap,  ix 

Right  Circular  Cone  (Fig.  299). — The  x  axis  contains  the  mass-center,  and 
is  parallel  to  the  base;  M  =  ^  irr'^ad. 

Frustum  of  a  Cone.  —  Let  R  =  radius  of  larger  base,  r  =  radius  of  smaller 
base,  and  a  =  altitude.     For  the  axis  of  the  frustum 

sphere.  —  Let  r  =  radius.     For  a  diameter 

k^=ir^;     I^^Sirr^b. 

■z 


Fig.  299  Fig.  300 

Hollow  Sphere.  —  Let  R  =  outer  and  r  =  inner  radius.     For  a  diameter 
yfe2=  |(^5_^)_^  (/23_^).     I  ^  ^^^Tr{R'-r>)8. 

Ellipsoid.  —  Let  2  a,  2  b,  and  2  c  =  length  of  axes.  For  the  axis  whose 
length  =  2  c, 

k^  =  l  (a2  _|_  j2) .     /  =  ^^  Trahcb  (a2  +  b""). 

Paraboloid  Generated  by  Revolving  a  Parabola  about  its  Axis.  —  Let  r  = 
radius  of  base  and  h  =  its  height.     For  the  axis  of  revolution 

Ring  (Fig.  300).  — The  x  axis  contains  the  mass-center  and  is  parallel  to  the 
plane  of  the  ring;   the  y  axis  is  the  axis  of  the  ring. 

K''-hR'+l  r-;        h  =  Tf'Rr'd  (R'  +  I  r'). 

ky'  =   7^2  _|_  I  ^.2.  7^   ^   2  ^2J^r2^  (7^2   _^  I  ^2), 

§  4.  Experimental  Determination  of  Moment  of  Inertia.  —  When 
the  body  is  so  irregular  in  shape  that  the  moment  of  inertia  desired  cannot 
be  computed  easily,  then  an  experimental  method  may  be  simpler.  There 
are  several  experimental  methods  available. 

By  Gravity  Pendulum.  —  This  method  is  available  if  the  body  can  be 
suspended  and  oscillated,  like  a  pendulum,  about  an  axis  coinciding  with  or 
parallel  to  the  line  with  respect  to  which  the  moment  of  inertia  is  desired. 
Let  T  =  the  time  of  one  complete  (to  and  fro)  oscillation,  c  =  distance  from 
the  mass-center  to  the  axis  of  suspension,  W  =  weight  of  the  pendulum,  g  = 
acceleration  due  to  gravity,  k  =  radius  of  gyration,  and  I  =  moment  of 
inertia  about  the  axis  of  suspension;  then 

T      —                              VcW 
k= —  Vcg         and        /  = 5-  •  (i) 

2  TT  4  TT'^ 


Art.  36  ^75 

Above  formulas  are  based  on  the  formula  for  the  time  of  oscillation  or 
period  of  a  pendulum  T  =  2tv  Vk^cg  (see  Art.  39).  If  the  axis  of  suspen- 
sion does  not  coincide  with  the  Hne  about  which  the  moment  of  inertia  is 
desired,  then  it  remains  to  ''transfer"  /  to  that  Hne  (see  §  2). 

The  desired  moment  of  inertia  can  be  determined  without  any  time  obser- 
vation as  follows:  From  the  same  axis  about  which  the  suspended  body 
oscillates  suspend  a  "  mathematical  pendulum,"  a  very  small  bob  with  cord 
suspension  (see  Art.  39) ;  adjust  the  length  of  the  cord  so  that  the  periods 
(times  of  oscillation)  of  bob  and  body  become  equal;  then 

k  =  Vd,  and  /  =  Wd/g,  .  (2) 

where  /  =  the  distance  from  the  center  of  the  bob  to  the  axis  of  suspension 
and  k,  W,  c,  I  have  the  same  meaning  as  above.  The  foregoing  result  is 
based  on  the  fact  that  k^/c  (for  the  pendulum)  equals  the  length  /  of  the 
mathematical  pendulum  (see  Art.  39). 

By  Torsion  Potdulum.  —  The  torsion  pendulum  here  referred  to  would 
consist  of  an  elastic  wire  suspended  in  a  vertical  position,  the  lower  end 
being  fashioned  or  terminated  in  a  disk  so  that  objects,  whose  moments  of 
inertia  are  to  be  determined,  may  be  suspended  on  the  wire  and  made  to 
oscillate  about  its  axis.  Let  /  =  the  (observed)  period  (time  of  one  oscilla- 
tion) of  the  bare  pendulum,  h  =  the  (observed)  period  of  the  pendulum  when 
it  is  loaded  with  a  body  A  which  is  so  regular  i.\  shape  (as  a  cube  or  cyUnder) 
that  its  moment  of  inertia  about  the  axis  of  oscillation  can  be  computed 
easily,  and  h  =  the  (observed)  period  of  the  pendulum  when  it  is  loaded  with 
the  body  B  whose  moment  of  inertia  is  desired;  further  let  h  =  the  (com- 
puted) moment  of  inertia  of  A  and  I2  =  the  m.oment  B  about  the  axis  of 
suspension.  B  should  be  suspended  so  that  the  axis  of  suspension  coincides 
with  or  is  parallel  to  the  line  (of  B)  about  which  the  moment  of  inertia  is 
desired.     Then 

l2=h{k-t)^{h-t).  (3) 

This  result  is  based  on  the  fact  that  the  square  of  the  period  of  a  torsion 
pendulum  is  proportional  to  the  moment  of  inertia  of  the  pendulum  with 
respect  to  the  axis  of  oscillation.  Thus,  if  /  =  the  moment  of  inertia  of  the 
bare  pendulum,  and  C  the  proportionality  factor,  then 

f  =  CI,  k^  =  C  (/  +  /i),  and  /o-  =  C  (/  +  A). 

These  three  equations  may  be  combined  so  as  to  eliminate  C  and  I  and  thus 
give  equation  (3). 

If  B  cannot  be  suspended  so  as  to  make  the  axis  of  oscillation  and  the  line 
(of  B)  about  which  the  moment  of  inertia  of  B  is  desired  coincident,  then  it 
remains  to  reduce,  or  transform,  h  to  that  line  (see  §  20  of  this  article). 


iy6  Chap,  ix 

37.   Rotation 

§  I.  A  rotation  is  such  a  motion  of  a  rigid  body  that  one  line  of  the  body  or 
of  the  extension  of  the  body  remains  fixed.  The  fixed  line  is  the  axis  of  the 
rotation.  The  motion  of  the  flywheel  of  a  stationary  engine  is  one  of  rotatiorv^ 
and  the  axis  of  rotation  is  the  axis  of  the  shaft  on  which  the  wheel  is  mounted; 
the  motion  of  an  ordinary  clock  pendulum  is  one  of  rotation,  and  the  axis  of 
rotation  is  the  horizontal  line  through  the  point  of  support  and  perpendicular 
to  the  axis  of  the  pendulum.  Obviously  all  points  of  a  rotating  body,  except 
those  on  the  axis  if  any,  describe  circles  whose  centers  are  in  the  axis  and  whose 
planes  are  perpendicular  to  the  axis.  The  plane  in  which  the  mass-center  of 
the  body  moves  will  be  called  the  plane  of  the  rotation,  and  the  intersection  of 
the  axis  of  rotation  and  the  plane  of  rotation  will  be  called  center  of  rotation.  All 
points  of  the  body  on  any  line  parallel  to  the  axis  move  alike;  hence  the  motion 
of  the  projection  of  the  line  on  the  plane  of  the  motion  represents  that  of  all 
the  points,  and  the  motion  of  the  body  itself  is  represented  by  the  motion  of 
its  projection. 

By  angular  displacement  of  a  rotating  body  during  any  time  interval  is 
meant  the  angle  described  during  that  interval  by  any  line  of  the  body  perpen- 
dicular to  the  axis  of  rotation.  Obviously  all  such  lines  de- 
scribe equal  angles  in  the  same  interval,  and  we  select  a  line 
which  cuts  the  axis.  Let  the  irregular  outline  (Fig.  301)  rep- 
resent a  rotating  body,  the  plane  of  rotation  being  that  of 
the  paper,  and  0  the  center  of  rotation.  Let  P  be  any  point 
and  6  the  angle  XOP,  OX  being  any  fixed  line  of  reference. 
Fig.  301  ^g  customarily,  6  is  regarded  as  positive  or  negative  according 

as  OX  when  turned  about  0  toward  OP  moves  counter  clockwise  or  clockwise. 
If  di  and  6i  denote  initial  and  final  values  of  6  corresponding  to  any  rotation, 
then  the  angular  displacement  =  Oo  —  6\. 

The  angular  velocity  of  a  rotating  body  is  the  time-rate  at  which  its  angular 
displacement  occurs;  or,  otherwise  stated,  it  is  the  time-rate  at  which  any  line 
of  the  body  perpendicular  to  the  axis  describes  angle.  The  time-rate  at  which 
OP  describes  angle,  or  the  time-rate  (of  change)  of  6  is  dd/dt  (see  Art.  28, 
Note).     Hence,  if  co  denotes  angular  velocity, 

CO  =  de/dt.  (i) 

Any  angular  displacement  divided  by  the  duration  of  that  displacement  gives 
the  average  angular  velocity  for  that  duration  or  interval  of  time.  If  the  body 
is  rotating  uniformly  (describing  equal  angles  in  all  equal  intervals  of  time), 
then  the  average  velocity  is  also  the  actual  velocity. 

The  formulas  for  angular  velocity  imply  as  wm/*  the  angular  velocity  of  a 
body  rotating  uniformly  and  making  a  unit  angular  displacement  in  each  unit 
time.  There  are  several  such  units;  thus,  one  revolution  per  minute,  one 
degree  per  hour,  one  radian  per  second,  etc.     The  last  is  the  one  usually  used 

*  For  dimensions  of  units  of  angular  velocity  and  acceleration,  see  Appendix  A, 


Art.  37  177 

herein.  An  angular  velocity  must  be  regarded  as  having  sign,  the  same  as  that 
of  dd/dt.  Since  dd/dt  is  positive  or  negative  according  as  d  increases  or  de- 
creases algebraically,  the  angular  velocity  of  a  rotating  body  at  any  instant 
is  positive  or  negative  according  as  it  is  turning  in  the  counter  clockwise  or 
clockwise  direction  at  that  time. 

The  angular  acceleration  of  a  rotating  body  is  the  time-rate  (of  change)  of 
its  angular  velocity.  If,  as  in  the  preceding,  co  denotes  the  angular  velocity, 
then  the  general  expression  for  the  time-rate  of  the  angular  velocity  is  do:/dt; 
hence  if  a  denotes  the  angular  acceleration, 

a  =  dio/di  =  d^d/dtK  (2) 

The  change  in  angular  velocity  which  takes  place  during  any  interval  of  time 
divided  by  the  length  of  the  interval  gives  the  average  angular  acceleration  for 
that  interval.  If  the  velocity  changes  uniformly,  then  this  average  accelera- 
tion is  also  the  actual  acceleration. 

The  foregoing  formulas  imply  as  unit  *  the  angular  acceleration  of  a  body 
whose  angular  velocity  is  changing  uniformly  and  so  that  unit  angular  velocity- 
change  occurs  in  each  unit  time.  One  revolution  per  second  per  second,  one 
radian  per  second  per  second,  etc.,  are  such  units.  An  angular  acceleration 
must  be  regarded  as  having  sign,  —  the  same  as  that  of  dw/dt.  Since  dw/dt 
is  positive  or  negative  according  as  co  increases  or  decreases  algebraically,  an 
angular  acceleration  is  positive  or  negative  according  as  the  angular  velocity 
is  increasing  or  decreasing  (algebraically). 

There  are  simple  relations  between  the  linear  velocity  v  and  linear  acceleration 

a  of  any  point  P  of  a  rotating  body  and  the  angular  velocity  and  acceleration  of  the 

body.    Let  r  —  the  distance  of  P  from  the  axis  of  rotation,  s  =  distance  travelled 

by  P  in  any  time  from  some  fixed  point  in  the  path  of  P,  and  6  —  the  angle 

described  by  the  radius  to  P  in  that  same  time.     Then  s  =^  rd  ii  dhe  expressed 

in  radians;  ds/dt  —  r  dd/dt,  or 

V  —  roi. 

Differentiating  again,  we  find  that  dv/dt  =  r  do^/dt,  or 

at  =  ra;     also     «„  (=  v'^/r)  =  rco^. 

Here  at  and  a„  mean  the  tangential  and  normal  components  of  the  acceleration 
of  P  (Art.  34). 

§  2.  Equation  of  Motion.  —  We  have  already  called  attention  to  the 
fact  (Art.  36,  footnote)  that  in  the  case  of  rotation  the  angular  acceleration 
is  proportional  to  the  algebraic  sum  of  the  moments  of  all  the  external  forces 
acting  on  the  body  directly  and  to  the  moment  of  inertia  of  the  body  inversely, 
both  moments  being  about  the  axis  of  rotation.  Or,  if  To  and  Iq  be  used  to 
denote  these  moments,  and  a  =  the  angular  acceleration,  then  a  is  proportional 
to  {Tq  -^  Iq)  ;  and,  if  systematic  units  (Art.  4)  be  used  then 

To  =  ha  =  Mko'a,  (3) 

*  For  dimensions  of  units  of  angular  velocity  and  acceleration,  see  Appendix  A. 


178  Chap,  ix 

where  M  =  mass  of  the  body  and  ko  =  its  radius  of  gyration  about  the  axis  of 
rotation.  If  W/g  be  written  for  M  (Art.  4,  §  2),  then  any  unit  of  force  (in- 
cluding W),  any  unit  of  length,  and  any  unit  of  time  may  be  used  in  (3). 

The  foregoing  is  called  the  equation  of  motion  for  a  rotation;  it  may  be  de- 
rived from  a  consideration  of  the  torque,  about  the  axis  of  rotation,  of  all 
the  forces  acting  on  each  particle  of  the  body.  Let  P'  (Fig.  302) 
represent  a  particle  of  the  rotating  body  not  shown,  m'  =  its 
mass,  and  a'  =  its  acceleration.  Then  the  resultant  of  all 
the  forces  acting  on  P'  =  m'a' ,  and  the  tangential,  normal, 
and  axial  components  of  this  force  are  m'o/,  m'an  ,  and  o 
respectively.  Similarly  the  tangential,  normal,  and  axial 
components  of  the  resultant  of  all  the  forces  acting  on  the 
Fig.  302  second  particle  P"  are  m"at",  m"an",  and  o.  All  the  radial 
or  normal  components  are  directed  toward  the  axis  of  rotation,  and  all  the 
tangential  components  clockwise  or  counter  clockwise.  Now  the  torque  of  all 
the  forces  acting  on  P'  equals  the  torque  of  m'at  and  m'an  ;  this  torque  = 
m'at'r' .  Similarly  the  torque  of  all  the  forces  acting  on  P"  =  m"ai"r" . 
Hence  the  torque  of  the  forces  acting  on  all  the  particles  equals 

m'at'r'  +  m"at"r"  +    •    •    •   =  m'r'ar'  +  m"r"ar"  +    •    •    •    =  cvSwr^  =  ah. 

Now  the  system  of  forces  acting  on  all  the  particles  consists  of  internal  and 
external  forces.  The  internal  forces  jointly  have  no  torque  since  they  consist 
of  pairs  of  coUnear,  equal,  and  opposite  forces.  Hence,  the  torque  of  the  ex- 
ternal forces  equals  Ua. 

Examples.  —  i .  Fig.  303  represents  a  circular  disk  of  cast  iron  4  inches  thick 
and  3  feet  in  diameter.  It  is  supported  on  a  fixed  horizontal  shaft  3  inches  in 
diameter.  A  cord  is  wrapped  around  the  disk,  and  then  a  pull  P  =  ^...j'_.^ 
100  pounds  is  applied  to  the  cord  as  shown.  What  is  the  angular 
acceleration  of  the  disk?  The  external  forces  acting  on  the  disk  and 
cord  are  the  weight  of  the  disk  and  cord  P,  and  the  reaction  of  the 
shaft.  Only  one  of  these,  P,  has  a  moment  about  the  axis  of  rota- 
tion. We  are  assuming  that  the  disk  is  homogeneous  so  that  the 
center  of  gravity  is  in  the  axis  of  rotation,  and  that  the  shaft  is  '  '^°"^ 
frictionless.  Tq  of  equation  (3)  is  therefore  100  X  1.5  =  150  foot-pounds. 
Now  the  square  of  the  radius  of  gyration  of  the  disk  about  the  axis  of  rotation 
is  I  (1.52  +  0.125^)  =  1. 133  feet^  (Art.  36).  And  since  the  weight  of  the  disk 
is  .T053  pounds,  its  moment  of  inertia  about  the  axis  of  rotation  is  (1053  -^  32.2) 
1.J33  =  37-0  slug-feet^.  Hence  the  angular  acceleration  of  the  disk  is  150  -^ 
37.0  =  4.0  radians  per  second  per  second. 

2.  Suppose  that  a  turning  force  P  in  the  preceding  example  is  supplied  not 
"by  hand  "  but  by  means  of  a  body  suspended  from  the  cord,  and  suppose 
that  the  body  weighs  100  pounds.  Obviously  the  system  (disk  and  suspended 
body)  moves  with  acceleration ;  hence  the  two  forces  acting  on  the  body  (gravity 
and  the  pull  P  of  the  cord)  are  not  equal  or  balanced  but  have  a  resultant 


Art.  37 


179 


downward  (direction  of  the  acceleration  of  the  body).  That  resultant  is 
100  —  P,  and  it  equals  the  product  of  the  mass  and  acceleration  of  the  body, 
or  100  —  P  =  (100  -V-  32.2)  X  a  where  a  =  the  acceleration.  The  torque  on 
the  disk  is  P  X  i-S,  and  1.5  P  =  /a  =  37.0  a.  But  a  =  the  tangential 
acceleration  of  any  point  on  the  rim  of  the  disk  =  1.5  X  a,  or  a  =  1.5  a. 
These  three  equations 

100  -  P  =   (100/32.2)  a,     1.5  P  =  37.0  a,     and     a  =1.5  a, 

solved  simultaneously  give  a  =  3.41  radians  per  second  per  second,  less  than 
in  example  i  as  was  to  be  expected,  because  the  pull  P  in  this  example  is  less 
than  100  pounds.  The  value  of  P  as  obtained  from  the  foregoing  equations 
is  84.1  pounds. 

3.  In  Fig.  304  we  take  weight  of  A  =  64  pounds,  of  P  =  96  pounds,  and  of 
pulley  C  =  144  pounds;  assume  coefficient  of  friction  under  B  =  I  ior  sliding, 
axle  friction  zero;  take  diameter  of  pulley  =  2  feet  6  inches,  and  the  radius 
of  gyration  of  the  pulley  about  the  axis  of  rotation  =  10.6  inches.  We  show 
how  to  determine  the  acceleration  of  the  system.  Let  a  =  acceleration  of  A 
and  B,  and  a  =  (angular)  acceleration  of  the  pulley.  Obviously  a  =  1.25  a. 
Let  us  now  consider  the  forces  acting  on  each  body  A,  B,  and  C.  On  A  there 
are  two,  —  gravity  (64  pounds)  and  the  pull  of  the  cord  Pi  (see  Fig.  305).     On 


Fig.  304 


96 /bs. 


B 


¥ 


F 

Fig.  306 


Fig.  307 


B  there  are  three,  —  gravity  (96  pounds),  the  pull  of  the  cord  P2,  and  the  re- 
action of  the  supporting  surface  D  (see  Fig.  306  where  this  latter  force  is 
represented  by  two  components  N  and  F).  On  the  pulley  there  are  three 
forces,  —  gravity  (144  pounds),  the  reaction  Q  of  the  axle,  and  the  pressure  of 
the  cord.  Since  the  mass  of.  the  cord  is  negligible,  the  tension  at  any  point  of 
the  cord  from  A  to  the  pulley  is  Pi,  and  at  any  point  from  B  to  the  pulley  it  is 
P2.  Hence  the  pressure  of  the  cord  against  the  pulley  equals  the  resultant  of 
Pi  and  P2  (Fig.  307),  and  that  pressure  is  equivalent  to  Pi  and  P2.  Therefore 
the  equation  of  motion  becomes  (Pi  -  P2)  1.25  =  (i44  ^  32)  (10.6  -^  i2ya 
=  4.5  X  0.778  X  a  =  3.5  a.  Since  the  acceleration  of  B  is  toward  the  right, 
the  resultant  force  on  it  acts  in  that  direction  and  equals  P2  -  P  =  P2  -  ^  N 
=  P2  -  ^,  96  =  Po  -  19.2;  and  hence  P2  -  19-2  =  (96  "^  32)  <i  =  3  «•  Since 
the  acceleration  of  A  is  downward  the  resultant  force  on  A  acts  in  that  direc- 
tion and  equals  64  -  Pi;  hence  64  -  Pi  =  (64  -^  32)  a  =  2  a.  Now  solving 
the  three  equations  of  motion, 

(Pi  -  ^^2)  X  1.25  =  3.5  a,     P2  -  19-2  =  3  a>     and     64  -  Pi  =  2  a, 


l8o  Chap,  ix 

together  with  a  =  1.25  a,  we  j&nd  that  a  =  6.19  feet  per  second  per  second, 
and  a  =  4.95  radians  per  second  per  second.  The  equations  also  show  that 
Pi  =  51.62  pounds,  and  P2  =  37.77  pounds. 

38.   Axle  Reactions 

§  I.  Simple  Cases.  —  Rotating  bodies  are  commonly  supported  by  shafts 
upon  or  with  which  the  bodies  rotate.  In  such  a  case,  axle  reaction  means 
the  force  which  the  shaft  exerts  upon  the  rotating  body.  To  determine 
such  a  force  we  make  use  of  the  principle  of  the  motion  of  the  mass-center. 
The  principle  states  (Art.  34)  that  the  algebraic  sum  of  the  components  —  along 
any  line  —  of  all  the  external  forces  acting  on  a  body,  moving  in  any  way, 
equals  the  product  of  the  mass  of  the  body  and  the  component  of  the  accelera- 
tion of  the  mass-center  along  that  line.  In  general,  the  principle  furnishes 
three  independent  equations,  one  for  each  of  three  rectangular  lines  of  resolu- 
tion. If  the  mass-center  of  the  (rotating)  body  does  not  lie  in  the  axis  of  rota- 
tion then  there  are  three  lines  of  resolution  which  are  generally  more  convenient 
to  use  than  any  others,  and  these  we  now  describe.  Let  the  circle  (Fig.  30S) 
be  the  path  of  the  mass-center  of  a  rotating  body  (not 
shown),  O  be  the  center  of  rotation  (intersection  of  the  axis 
of  rotation  and  plane  of  the  path  of  the  mass-center),  and 
C  be  the  mass-center.  Then  the  three  convenient  lines  are 
the  axis  of  rotation,  the  line  OC,  and  a  line  perpendicular  to 
the  first  two.  The  directions  of  these  lines  are  called  re- 
FiG.  30S  spectively  axial,  radial  or  normal  (OC  being  a  radius  and 

normal  of  the  circle),  and  tangential  (the  third  line  being  parallel  to  the  tan- 
gent at  C).  Now  let  I1F<,I!F„,  and  ^Fa  =  the  algebraic  sums  of  the  tangen- 
tial, normal,  and  axial  components  of  all  the  external  forces  acting  on  any 
rotating  body;  at  and  a„  =  the  tangential  and  normal  components  of  the  ac- 
celeration of  its  mass-center  —  the  axial  component  of  the  acceleration  equals 
zero;  and  M  =  the  mass.     Then 

2F,  =  Alat,     :SF„  =  Man,     2F,  =  o.  (i) 

Systematic  units  (Art.  31)  must  be  used  in  the  foregoing.  If  W/g  be  substituted 
for  M  (Art.  31)  then  any  unit  may  be  used  for  force  (including  weight),  any 
unit  for  length,  and  any  unit  for  time. 

Let  r  =  radius  of  the  circle  described  by  the  mass-center,  v  =  velocity  of 
the  mass-center,  a  =  angular  acceleration,  and  co  =  angular  velocity  of  the 
rotating  body  at  the  instant  under  consideration;   then  (see  Art.  37,  §1) 

at  =  ra,     and     a„  —  v'^/r  =  rco^, 

and  we  may  use  these  in  equations  (i). 

If  the  mass-center  of  the  rotating  body  is  in  the  axis  of  rotation,  then  the 


Art.  38 


181 


Fig.  309 


acceleration  of  the  mass-center  is  always  zero,  and  the  algebraic  sum  of  the 
components  of  the  external  forces  along  any  line  equals  zero. 

Examples.  —  i.  AB  (Fig.  309)  is  a  bar  of  wrought  iron  1.5  inches  (perpen- 
dicular to  paper)  X  4  inches  X  6  feet,  suspended  from  a  horizontal  axis  at  A. 
Suppose  that  the  bar  is  made  to  rotate  and  is  then  left  to  it-  r, 
self  rotating  under  the  influence  of  gravity,  the  axle  reaction, 
and  the  initial  velocity  given  to  it.  Suppose  further  that 
the  initial  velocity  was  such  that  when  the  bar  gets  into  the 
position  shown,  the  angular  velocity  is  60  revolutions  per 
minute.  Required  the  axle  reaction  in  the  position  shown. 
The  only  forces  acting  on  the  body  are  its  weight  W  =  120 
pounds,  and  the  axle  reaction  represented  by  two  components 
Ri  and  R^.  We  neglect  the  axle  friction;  then  the  lines  of 
actions  of  Ri  and  R-i  cut  the  axis  of  rotation,  and  the  equation  of  motion  (Art. 
37)  becomes  IF  (2  sin  35°)  =  la.  Now  I  =  (IF/32.2)  ^2  =  (120/32.2)  7.01; 
hence  a  =  5.26  radians  per  second  per  second,  and  Cj  =  2  X  5.26  =  10.52 
feet  per  second  per  second.  The  angular  velocity,  60  revolutions  per  minute, 
equals  6.28  radians  per  second;  hence  a„  =  2  X  6.28^  =  78.8  feet  per  second 
per  second.     Finally,  equations  (i)  become 

120  sin  35°  —  Ri=  (120/32.2)  10.52  =  39.2,  and 
R2  —  120  cos  35°  =  (120/32.2)  78.8  =  294. 

From  the  first  Ri  =  29.7,  and  from  the  second  R2  =  392  pounds. 

2.  AB  (Fig.  310)  is  a  simple  brake  for  retarding  the  motion  of  the  drum  C 
and  suspended  body  IF.  Let  IF  =  2000  pounds,  weight  of  the  drum  =  1800 
pounds,  radius  of  gyration  of  drum  about  axis  of  rotation  =2.5  feet,  coeflBcient  of 


Fig.  310 


friction  "  between  "  brake  and  drum  =  0.5.  Suppose  that  IF  is  descending  and 
the  brake  pull  P  is  1000  pounds.  Required  the  axle  reaction  on  the  drum.  Fig. 
311  shows  all  the  forces  acting  on  the  drum,  — its  own  weight  (1800  poimds), 
the  brake  pressure  represented  by  two  components  N  (normal  pressure)  and 
F  (friction),  the  pull  T  of  the  rope,  and  the  axle  reaction  represented  by  two 
components  Ri  and  R2.  From  a  consideration  of  the  forces  acting  on  the  brake 
it  is  plain  that  N  =  (1000  X  6.5)  -^  1.5  =  4333  pounds;  and  hence  F  =  0.5 
X  4333  =  2167  pounds.  Now  in  order  to  get  T  we  write  out  the  equations 
of  motion  of  the  drum  and  the  suspended  body.     Since  F  is  greater  than  the 


l82 


Chap,  ix 


weight  of  the  body  the  velocities  of  drum  and  body  are  being  decreased;  hence 
T  is  greater  than  W  but  less  than  F.  If  a  =  the  acceleration  of  the  drum  and 
a  =  the  acceleration  of  the  suspended  body,  then  the  equations  of  motion  are 

2167  y.  2)  ~  T^  ^  i  ^  (1800/32.2)  2.52a,     and 
T  —  2000  =  (2000/32.2)  a. 

These  equations  and  a  =  T)^^^  solved  simultaneously,  give  T  =  2103  pounds. 
Since  the  acceleration  of  the  mass-center  of  the  drum  equals  zero, 

2167  —  Ri^  o,     or     Ri  =  2167,     and 
R2  —  4333  —  1800  —  2103  =  o,     or    R2  =  8236  pounds. 

Therefore  the  axle  reaction  =  V{2i6'j'^  +  8237')  =  8500  pounds  inclined  up- 
wards and  to  the  left  at  an  angle  of  14I  degrees  with  the  vertical. 

§  2.  Non-simple  Cases.  —  Axle  reactions  cannot  be  determined  by  means 
of  equations  (i)  in  some  cases;  moment  equations  must  be  resorted  to.  The 
moment  equation  To  =  ha  (Art.  37)  is  available  for  all  cases  but  additional 
ones  may  be  needed.  It  will  be  recalled  that  To  is  the  torque  of  all  the  external 
forces  about  the  axis  of  rotation;  we  will  presently  deduce  expressions  for 
the  torque  about  other  lines,  —  first  for  a  body  of  any  shape  and  then  for 
bodies  of  certain  common,  symmetrical  shapes. 

Body  of  any  Shape.  —  Let  the  axis  of  rotation  be  taken  as  the  z  axis  of  an 
x-y-z  coordinate  frame,  the  x  axis  containing  the  mass-center  on  the  positive 


k>x4     I 


<  S    Q.  X 


I 

/     Mrcc 


Z      t|        '  symmetry  Z  'Mru)' 

Fig.  312  Fig.  313  Fig.  314  Fig.  315 

side  of  the  axis.     The  following  six  equations  apply  to  this  general  case: 


(i)    2F,  =  -  MFco2, 
(2)    llFy  =  M?a, 


T,=  -a  fzx  dM  +  aj2  Cyz  dM,        (4) 

Ty=  -  a  I  yzdM  -or  \  zx  dM,        (5) 


(3)    2F.  =  0,  T,=  ha  =  Mk'^a.  (6) 

Equations  (i),  (2),  and  (3)  follow  from  Art.  38  (page  180);  (6)  is  equation  (3) 
of  Art.  37,  the  notation  being  changed  to  agree  more  appropriately  with 
Fig.  312;   (4)  and  (5)  will  be  deduced  immediately. 

The  method  used  for  arriving  at  (4)  and  (5)  is  like  that  used  to  get  (6). 


Art.  38  183 

The  moment  of  the  resultant  of  all  the  forces  acting  on  each  particle  is  com- 
puted; then  such  moments  for  all  the  particles  are  summed;  and  since  the 
internal  forces  contribute  nothing  to  it,  this  sum  is  the  value  of  the  torque  of 
the  external  forces  about  the  axis  of  moments  being  used.  Let  P  (Fig.  312) 
represent  any  particle  of  the  rotating  body,  r  the  radius  of  the  circle  described 
by  P,  e  the  varying  angle  PAB,  (x,  y,  z)  the  coordinates  of  P,  m  its  mass, 
and  a  its  acceleration.  The  resultant  of  all  the  forces  acting  on  the  particle 
P  equals  ma.  Components  of  this  resultant  along  and  perpendicular  to 
PA  equal  mrur  and  mra.  The  first  component  acts  in  the  (radial)  direction 
PA ,  and  the  second  in  the  direction  of  the  tangential  component  of  a.  The 
moments  of  the  resultant  ma  about  the  x  and  y  axes  respectively  equal  the 
sums  of  the  moments  of  its  components;   these  sums  are 

—  mra  cos  6s  +  mror  sin  6z  —  —  mazx  +  mcJ'yz, 
and  —  mra  sin  dz  —  mroi^  cos  6z  =  —  mayz  —  moihx. 

If  now  we  add  all  such  moments  about  the  x  axis,  and  then  those  about  the 
y  axis,  for  all  the  particles  comprising  the  body,  we  arrive  at 
—  aLmzx-{- orllmyz,     and     —allmyz  —  (j?'Lmzx\ 

and  these  reduce,  for  a  continuous  body,  to  the  right-hand  members  of  (4) 
and  (5). 

(i)  The  body  is  homogeneous,  has  a  plane  of  symmetry,  and  rotates  about  an 
axis  perpendicular  to  that  plane.     In  this  case,  (4)  and  (5)  become 

Tx  =  o    and     Ty  =  o.  (4')  (5') 

For:  —  The  xy  plane  (Fig.  312)  now  coincides  with  the  plane  of  symmetry; 
hence,  for  any  particle  whose  coordinates  are  (.v,  y,  2)  there  is  a  corresponding 
one  whose  coordinates  are  (x,  y,  —z).  And  because  of  the  (assumed)  homo- 
geneity of  the  body  the  masses  of  the  particles  may  be  taken  equal.  It  fol- 
lows that  Swsx  =  o  and  Hmyz  =  o  for  the  two  particles.  Therefore,  these 
summations  extended  to  all  the  pairs  of  particles  comprising  the  body  equal 
zero,  and  hence  Tx  =  Ty  =  o  SiS  stated. 

Let  Fig.  313  represent  the  plane  of  the  symmetry  section  of  the  body,  C 
the  mass-center,  0  the  center  of  rotation,  Q  a  point  on  OC  extended  so  that 
OQ  =  k-/r.  In  general,  the  resultant  of  the  external  forces  is  a  single  force 
acting  in  the  xy  plane,  and  through  Q.  The  components  of  the  resultant 
along  and  perpendicular  to  OC  are  indicated  in  the  figure;  the  first  always 
acts  from  Q  toward  0  and  the  second  in  the  direction  of  the  tangential  accel- 
eration of  C.  Proof  of  this  statement  is  left  for  the  student  to  supply.  The 
fact  that  the  described  components  satisfy  equations  (i),  (2),  (3),  (4'),  (5'), 
and  (6)  may  be  regarded  as  sufficient  proof. 

When  the  angular  velocity  is  constant  (a  =  6),  the  resultant  of  the  external 
forces  is  a  force  directed  along  the  radius  CO  and  in  that  direction;  its  value 
is  Mroi^.  When  the  mass-center  is  in  the  axis  of  rotation  (r  =  o),  the  re- 
sultant is  a  couple;  its  plane  is  perpendicular  to  the  axis  of  rotation  and  its 
moment  is  /««.     When  both  a  and  the  r  =  o,  the  resultant  is  nil. 


184  Chap-  IX 

(ii)  The  body  is  homogeneous,  has  a  line  of  symmetry,  and  rotates  about  an 
axis  parallel  to  that  line.    In  this  case,  (4)  and  (5)  become,  as  in  (i), 

Tx  =  o    and     T,  =  o  (4")  (5") 

For:  —  Let  Pi  and  P2  (Fig.  314)  be  any  symmetrical  pair  of  particles  (so  that 
the  line  P1P2  is  perpendicular  to  the  axis  of  symmetry  and  is  bisected  by  that 
axis).  Also  let  the  coordinates  of  Pi  and  P2  be  (xiyiZi)  and  (x2}'2Z2)  respectively. 
Evidently  these  coordinates  are  related  as  follows: 

h  (xi  +  X2)  =  r,    yi=  -  y2,     and    Zi  =  Z2. 

On  account  of  homogeneity,  the  masses,  nii  and  m^,  of  the  Pi  and  P2  may  be 
taken  equal;  hence  for  the  two  particles 

Hmyz  =  niiyiZi  +  m2y2Z2  =  wiyiZi  —  miyiZi  =  o, 
and  Smzx  =  miZiXi  +  ^W2Z2a-2  =  miZiXi  +  WiZi(2  r  —  Xi)  =  2  miZi?. 

It  follows  that,  if  the  summations  be  extended  to  all  the  pairs  of  particles 
constituting  the  body,  then 

fyz  dM  =  o,      and      jzx  dM  =  rjz  dM  =  rMz  (see  Art.  23), 

where  s  denotes  the  2  coordinate  of  the  mass-center;  but  the  mass-center  is 
in  the  XOY  plane,  and  hence  i  =  o.  Thus  finally,  we  see  from  (4),  (5), 
and  the  foregoing  results  that  T^,  and  T^  =  o  as  stated. 

All  the  remarks  under  case  (i)  about  the  resultant  of  the  external  forces 
hold  in  this  case. 

(iii)  The  body  is  homogeneous,  has  a  plane  of  symmetry,  and  rotates  about  an 
axis  in  that  plane.    In  this  case,  (4)  and  (5)  become 

r,  =  -  afzx  dM,    and     Ty  =  -  '^""^zx  dM.  (4'")  (5'") 

Yox:  —  The  xz  plane  coincides  with  the  plane  of  symmetry  in  this  case;  hence 
for  any  particle  whose  coordinates  are  (x,  y,  z)  there  is  a  corresponding  one 
whose  coordinates  are  {x,  -  y,  z).  It  follows  thatjjs  dM^o,  and,  from  (4) 
and  (5),  that  T^  and  Ty  have  the  values  stated  above. 

In  general,  the  resultant  of  the  external  forces  is  a  single  force  as  in  cases 
(i)  and  (ii);  but  in  this  case,  the  resultant  acts  not  in  the  xy  plane  but  in  a 
parallel  plane,  the  z  coordinate  of  which  is  jzxdM  ^  Mr.  See  Fig.  315; 
Q'  is  the  point  where  the  Une  of  action  of  the  resultant  pierces  the  plane  of 
symmetry.  Proof  of  the  foregoing  statements  is  left  for  the  student  to  supply. 
The  fact  that  the  described  resultant  satisfies  (i),  (2),  (3),  (4'"),  (s'"),  and 
(6)  may  be  regarded  as  sufficient  proof.* 

Centrifugal  Action  and  Dynamic  Balance.  —  It  is  common  experience  that 
a  rotating  body  unless  well  "balanced"  exerts  forces  upon  its  shaft  — and 
thus  upon  the  bearings  supporting  the  shaft  —  which  are  due  in  part  to  the 

*For  some  remarks  on  special  cases,  see  page  i88b. 


Art.  39 


185 


velocity  of  rotation.  Such  parts  or  components  of  the  forces  are  said  also  to 
be  due  to  the  "centrifugal  action  "  of  the  rotating  body;  and  the  components 
are  called  "  centrifugal  "  forces  or  pulls.  A  rotating  machine  part  so  shaped 
or  loaded  that  it  exerts  no  resultant  centrifugal  pull  on  its  shaft  or  bearhigs 
is  said  to  be  in  "running  "  or  "dynamic  balance." 

The  common  method  of  designing  for  running  balance  consists  in  arranging 
or  proportioning  the  various  (simple)  parts  of  the  rotating  body,  a  motor 
crank  shaft  for  example,  so  that  the  centrifugal  pulls  of  all  the  various  parts 
will  neutralize.*  Even  after  careful  design  and  manufacture  of  a  crank 
shaft,  say,  the  running  balance  may  be  imperfect.  Then  mechanical  methodsf 
are  resorted  to  for  completing  the  task. 

The  conditions  for  dynamic  balance  can  be  stated  with  reference  to  equa- 
tions (i)  to  (6):—  (a)  The  mass-center  of  the  body  must  be  in  the  axis  of 
rotation    {r=  o);    this   insures   "standing"   or   "static  balance."     (b)  The 

"products  of  inertia"  (see  Art.  57)  J  yzdM  &nd  jzxdM  must  equal  zero; 

this  with  (a)  insures  dynamic  balance.  For:— When  these  conditions  are 
fulfilled  and  the  angular  velocity  is  constant,  the  right-hand  members  of 
equations  (i)  to  (6)  equal  zero,  that  is  they  are  equilibrium  equations;  hence, 
the  bearing  pressures  are  independent  of  the  motion. | 

The  axis  of  rotation  of  a  body  which  is  dynamically  balanced  for  that 
axis  is  sometimes  called  a  "  free  axis  "  of  the  body;  for,  if  the  body  could  be 
rotated  about  that  axis  and  then  left  to  itself  entirely  free  from  all  external 
forces,  even  gravity,  it  would  continue  to  rotate  about  that  axis. 

39.  Pendulums 

§  I.  Gravity  Pendulum.  —  By  this  term  is  meant  the  common  pendulum, 
that  is  a  body  suspended  on  a  horizontal  axis  so  that  it  can  be  made  to  oscillate 
freely  under  the  influence  of  gravity.  A  real  pendulum  is  sometimes  called  a 
compound  or  physical  pendulum  to  distinguish  it  from  an  imaginary  one  con- 
sisting of  a  mass-point  or  particle  suspended  by  a  massless  cord;  this  latter  is 
called  a  simple  or  mathematical  pendulum.  Let  T  =  the  period  or  time  of  one 
complete  or  double  (to  and  fro)  oscillation,  k  =  the  radius  of  gyration  of  the 
pendulum  with  respect  to  the  axis  of  suspension,  c  =  distance  from  the  center 
of  gravity  of  the  pendulum  to  that  axis,  and  2  jS  =  the  angle  swept  out  by  the 
pendulum  in  one  single  oscillation.     Then,  as  will  be  shown  presently,  the 

period  is  given  closely  by  

T  =2T  Vk'/cg,  (i) 

*  For  a  good  treatment  of  dynamic  balancing,  see  the  book  by  Dunkerly  or  Sharpe  on 
Balancing  of  Engines. 

t  For  a  description  of  an  ingenious  balancing  machine,  Akimofif's,  see  American  Machinist 
for  May  18,  1Q16. 

X  For  a  theory  of  balancing  of  engines  based  on  these  conditions  extended  see  Lorenz, 
Tecknische  Mcchanik,  Band  I. 


1 86  Chap,  ix 

provided  that  j8  is  small.*  Since  jS  does  not  appear  in  this  formula  the  period 
of  any  pendulum  is  independent  of  /3;  that  is  all  small  oscillations  of  a  pendulum 
have  equal  periods  or,  as  we  say,  they  are  isochronous.  When  g  is  expressed  in 
feet  per  second  per  second  then  k  and  c  should  be  expressed  in  feet;  T  will  be 
in  seconds. 

For  the  derivation  of  equation  (i)  let  OG  (Fig.  316a)  be  a  pendulum  in  any 
swinging  position,  O  the  center  of  suspension,  G  the  center  of  gravity;  let  W  = 
the  weight  of  the  pendulum,  c  =  OG,  and  6  the  (varying)  angle  which  OG  makes 
with  the  vertical,  regarded  as  positive  when  the  pendulum  is  on  the  right  side 
of  the  vertical,  as  shown.  There  are  three  forces  acting  on  the  pendulum,  — 
gravity,  the  supporting  force  at  the  knife  edge,  and  the  pressure  of  the  sur- 
rounding air.  The  moment  of  the  first  force  about  the  axis  of  suspension  is 
Wc  sin  6;  the  moments  of  the  other  two  forces  we  take  as  negligible.  Hence 
the  resultant  torque  on  the  pendulum  in  any  position  =  Wc  sin  6  practically. 
The  angular  acceleration  =  <Pd/dfi  (see  Art.  37);  hence  according  to  equation 

(3)  of  that  article 

Wc  sin  6=  -  {W/g)  m^QldC-, 

the  negative  sign  being  introduced  because  sin  Q  and  d^d/df  are  always  oppo- 
site in  sign.     It  follows  readily  from  the  preceding  equation  that 

(Pe/df  =  -  (cg/k^)  sin  0  =  -A  sin  d, 

where  A  is  an  abbreviation  for  cg/k^.  We  will  assume  that  the  greatest  value 
of  6,  that  is  jS,  is  so  small  that  sin  d  and  6  are  nearly  equal;  then  as  a  good 
approximation  we  may  substitute  6  for  sin  6,  and  have 

deydf  =  -  Ad. 

To  integrate  this  simply,  let  u  —  dd/dt;  then  d^O/df  =  du/dt  =  (du/dd) 
{dd/dt)  =  {du/dd)  u,  and  hence 

(du/dO)  u  =  —  Ad,     or    udu  =  —  Ad  dd. 

Now  integrating  and  replacing  u  by  dd/dt,  we  get 

where  Ci  is  a  constant  of  integration.  Remembering  that  dd/dt  =  the  angular 
velocity  of  the  pendulum,  we  note  that  where  d  =  ^,  there  dd/dt  =  o;  there- 
fore for  these  (simultaneous)  values  the  preceding  equation  becomes  o  = 
-^AjS^-h  Ci,  or  Ci  =  ^  .4/32,  and  finally 

dd/dt  =  ±AWp^-d\ 

*  The  exact  value  of  the  period  is  given  by 

r  =  2,rVFAg[i  +  (iysin^^-}-(i.^ysin^^+  •  •  •  ]. 

If  /3  =  8  degrees  then  the  bracket  above  =  i. 001 2 2;  and  for  smaller  values  of  0  the  value 
of  the  bracket  is  still  nearer  unity.  Hence  the  error  in  the  approximate  formula  is  less 
than  one-eighth  of  one  per  cent  if  /3  does  not  exceed  8  degrees. 


Art.  39  187 

The  positive  sign  is  to  be  used  when  dd/dt  is  positive;  that  is  when  the  pen- 
dulum is  swinging  in  the  positive  direction.  Now  let  r  =  the  time  required 
for  the  pendulum  to  swing  out  from  its  lowest  to  its  highest  position  on  the 
right,  that  is  while  d  changes  from  o  to  /3.  To  get  a  value  of  this  time  we 
integrate  the  preceding  equation  as  follows: 

VA    I    dt^  +    I        ,  ,     or    r  =  V  T    sm-1  -      =  -  W  —  • 

Jj  Jo    \/,32  -  02 '  V  ,4  L         /3jo       2  V  eg 

Let  t'  =  the  time  required  for  a  swing  from  the  extreme  right  position  to  the 
lowest  position,  that  is  while  Q  changes  from  /3  to  o.  To  get  this  time  we 
integrate  as  follows: 

Jo  Ji3  V/32-02'  y  Ai     ^\p    2  V eg 

Hence  r  and  t'  are  equal,  as  was  to  be  expected.  Finally,  the  time  of  one 
complete  oscillation  =  47  =  2x  V^-/t:g,  as  was  to  be  shown. 

Let  k  =  the  radius  of  gyration  of  the  pendulum  with  respect  to  an  axis 
through  its  center  of  gravity  and  parallel  to  the  axis  of  suspension;  then 
k^  ^R  -\-  c-  (see  Art.  36),  and  hence 


gc  g  g\        c-J 


(2) 


Contrary  to  common  belief,  the  period  does  not  increase  for  all  increases  in  c, 
the  distance  from  the  center  of  gravity  to  the  axis  of  suspension.  For  examin- 
ing the  foregoing  expression  for  T  with  reference  to  a  variation  in  c,  we  find  that 

dT^       TV  {c"  -  ?) 
dc       cV[gc(c2-fP)]* 

Now  this  is  negative  for  all  values  of  c  less  than  k,  and  positive  for  all  values 
greater  than  k.  Hence  when  c  is  less  than  k  an  increase  in  c  decreases  T; 
when  c  is  greater  than  k  an  increase  in  c  increases  T.  When  c  =  k,  then 
dT/dc  =  o,  and  T  has  its  least  value  equal  to  2x\/(2  k/g). 

In  the  case  of  a  simple  pendulum  of  length  I,  the  radius  of  gyration  k  =  I 
and  also  c  =  I;  hence  the  period  of  a  simple  pendulum  is  given  by 

r=27rv77g.*  (3) 

A  physical  pendulum  and  a  simple  pendulum  whose  periods  are  equal  are  said 
to  be  equivalent.     Periods  are  equal  if  k"/cg  =  l/g,  or  /  =  k'/c. 

Imagine  the  entire  mass  of  a  (real)  pendulum  concentrated  into  a  point  Q 

*  Strictly  speaking,  a  simple  or  mathematical  pendulum  can  exist  only  in  imagination, 
but  a  real  pendulum  made  of  a  small  bob  suspended  by  means  of  a  cord  may  be  regarded  as  a 
simple  pendulum  in  computing  period.  That  is  the  period  of  the  cord-bob  pendulum  =  £t 
V77^  where  /  =  the  distance  from  the  axis  of  suspension  to  the  center  of  the  bob.  For  k/£ 
for  the  cord-bob  pendulum  is  small  compared  to  i,  and  hence  equation  (2)  gives  T  =  2w  ^c/g 
practically. 


j88 


Chap,  ix 


(Fig.  316a),  whose  distance  from  the  center  of  suspension  0  equals  k'^/c.  Then, 
as  just  shown,  the  period  of  such  an  (imaginary)  simple  or  mathematical 
pendulum  would  be  2TVQ/g),  where  /  is  the  length  OQ  or  k'^/c;  hence 
the  period  would  be  2  tt  Vik^cg),  that  is  equal  to  the  period  of  the  real 
pendulum. 

For  this  reason  Q  is  sometimes  called  the  center  of  oscillation  of  the  pen- 
dulum. (It  coincides  with  the  center  of  percussion,  see  Art.  48.)  The  dis- 
tance from  the  center  of  gravity  to  the  center  of  oscillation  is 


GQ  =  - 
c 


c  — 


c 


w] 


0, 


Fig.  316,  a,  b,  c 


"The  centers  of  suspension  and  of  oscillation  of  a 
pendulum  are  interchangeable,"  that  is  if  a  pendulum 
be  suspended  from  ()  (Fig.  316b),  then  0  becomes  the 
center  of  oscillation.  For,  suppose  that  Q'  is  the 
center  of  oscillation  corresponding  to  (2,  then 

hence  Q'  coincides  with  0.  It  follows  from  the  prop- 
erty of  interchangeability  that  the  periods  of  a  pendu- 
lum when  suspended  from  0  and  from  Q  are  equal. 

The  pendulum  is  our  best  device  for  accurately  determining  the  acceleration 
due  to  gravity  at  any  place.  We  have  only  to  determine  the  period  T  and 
the  length  k^jc  of  a  pendulum  at  the  place,  and  then  compute  g  from  the 
formula  T  =  2 -k  y/ {]r  I  eg) .  But  it  is  not  easy  to  determine  k^jc  directly. 
Captain  Kater  first  (18 18)  made  use  of  the  property  of  interchangeabiUty  of 
centers  of  suspension  and  oscillation  to  make  a  pendulum  whose  length  W-jc 
could  be  determined  accurately  and  readily.  Fig.  316c  represents  a  Kater 
pendulum  in  principle;  Oi  and  O2  are  two  knife  edges  as  shown  at  a  known 
distance  apart;  W  is  a  weight  which  can  be  slid  along  the  rod  and  clamped 
where  desired.  The  periods  of  oscillation  for  0\  suspension  and  O2  suspension 
would  be  different,  but  by  shifting  the  weight  and  trying  repeatedly,  the 
periods  can  be  made  equal.  When  the  periods  are  equal,  then  either  knife 
edge  is  the  axis  of  oscillation  for  the  other  as  axis  of  suspension,  and  O1O2 
(a  known  distance)  is  the  length  of  the  equivalent  simple  pendulum  or  the 
W'lc  of  the  formula.  By  means  of  a  Kater  pendulum  the  value  of  g  for  Wash- 
ington was  determined  to  be  980.100  centimeters  per  second  per  second.  Values 
of  g  at  many  other  places  have  been  determined  more  simply  —  by  comparing 
the  periods  of  oscillation  of  a  more  ordinary  pendulum  at  Washington  and  the 
other  places.  This  comparison  is  based  on  the  principle  that  the  squares  of 
the  periods  of  oscillation  of  any  pendulum  at  two  different  places  are  inversely 
proportional  to  the  values  of  g  at  those  places ;  hence  if  Ty,  and  T  =  the  periods 
at  Washington  and  some  other  station  and  g  =  the  acceleration  at  the  latter 
place,  then  g  =  980.1  {Tw/Tf- 


Art.  39  ,  1 88a 

§  2.  Torsion  Pendulum.  —  This  consists  of  a  heavy  bob  suspended 
vertically  by  means  of  a  Ught  elastic  wire,  the  wire  being  firmly  embedded 
in  the  bob  and  in  its  support.  Any  horizontal  couple  appUed  to  the  bob  will 
turn  the  bob  and  twist  the  wire.  If  the  couple  is  not  too  large  —  so  as  not 
to  stress  the  wire  beyond  its  "elastic  limit"  —  then  the  angular  displace- 
ment of  the  bob  will  be  proportional  to  the  moment  of  the  couple  applied. 
That  is,  if  C  and  C  =  the  moments  of  two  couples  apphed  successively  and 
6  and  6'  are  the  corresponding  angular  displacements  produced  by  the  couples, 
then  6/6'  =  C/C.  Hence,  the  moment  C  required  to  produce  any  displace- 
ment is  given  by  C  =  {C'/d')  6.  In  any  displaced  position  of  the  bob,  the 
wire  exerts  a  couple  on  the  bob  equal  to  the  appUed  couple. 

If  the  bob  were  released  from  any  position  of  (moderate)  angular  displace- 
ment j8,  it  would  oscillate  under  the  influence  of  the  couple  exerted  by  the 
wire.  We  will  assume  that  this  (varying)  couple  follows  the  law  expressed 
above.  Then  the  equation  of  motion  (rotation)  for  the  bob  would  be  (see 
equation  3,  Art.  37)  C  =^  la,  where  /  =  moment  of  inertia  of  the  bob  with 
respect  to  the  axis  of  the  wire  and  a.  =  the  (varying)  angular  acceleration. 
Since  a  =  drd/df,  and  6  and  d"d/dt"  are  opposite  in  sign,  the  equation  can  be 
written 

where  B  is  an  abbreviation  for  {C'/d')  -^  /.  This  last  equation  is  just  like  the 
equation  d^6/dt-  =  —  .40  of  §  i,  relating  to  the  motion  of  a  gravity  pendulum 
except  that  B  appears  in  one  equation  and  A  in  the  other.  Hence  the  form- 
ulas in  §  I  apply  to  this  section  if  A  be  changed  to  B.  Thus  the  time  of  one 
quarter  complete  oscillation  of  a  torsion  pendulum  is 


TT  = 


\/|[-iI=^v/c^ 


C/6' 

The  period  (one  complete  to  and  fro  oscillation)  equals  4  tt,  or 

T=  2ir\/l^{C'/d').  ^  (i) 

/  =  Mk^  =  {W/g)  k'  where  W  =  weight  and  k  =  radius  of  gyration  of  the 
bob  with  respect  to  the  axis  of  the  wire.  If,  in  a  numerical  case,  W  is  taken  in 
pounds  and  g  in  feet  per  second  per  second,  k  should  be  expressed  in  feet,  C 
in  foot-pounds,  and  6'  (always)  in  radians;  T  will  be  in  seconds.  C'/d'  (the 
ratio  of  the  moment  of  any  twisting  couple  to  the  angle  of  twist  produced) 
is  a  measure  of  the  torsional  stiffness  of  the  wire,  for  that  ratio  is  the  moment 
required  for  twisting  per  radian  of  twist.  Formula  (i)  shows  that  the  stiffer 
the  wire  the  less  the  period. 


i88b  Chap,  ix 

Continued  from  page  184. 
When  the  angular  velocity  is  constant  (a  =  o),  the  resultant  of  the  external 
forces  is  a  single  force  R  {  =  Mro)^)  acting  as  shown  in  Fig.  315  and  its  dis- 
tance from  CO  is 


/ 


zxdM  -T-  Mr. 


When  the  mass-centre  is  in  the  axis  of  rotation  {r  =  o),  the  resultant  is  a 
couple  whose  moments  about  the  x,  y,  and  0  axes  respectively  are 


—  ai  zxdM,  —  <a~  j  zxdM,  and  Mk^a. 


When  both  a  and  r  =  o,  the  resultant  is  a  couple  whose  moments  about 
the  X,  y,  and  z  axes  respectively,  are 


o,  —  ar  j  zxdM,  and  o. 


CHAPTER  X 

WORK,  ENERGY,  POWER 
40.   Work 

§  I.  Definitions,  Etc.  —  Work  is  a  common  word  and  has  many  mean- 
ings (see  dictionary),  but  it  is  used  in  a  single  special  sense  in  Mechanics. 
Work  is  said  to  be  done  upon  a  body  by  a  force  —  also  by  the  agent  exerting 
the  force  —  when  the  point  of  application  of  the  force  moves  so  that  the  force 
has  a  component  along  the  path  of  the  point  of  application.  This  component 
will  be  called  the  working  component  of  the  force;  and  the  length  of  the  path 
of  the  point  of  application  the  distance  through  which  the  force  acts.  The 
amount  of  work  done  by  the  force  is  taken  as  equal  to  the  product  of  the 
working  component  and  the  distance  through  which  the  force  acts. 

The  meaning  of  this  measure  of  work  done  by  a  force  is  clear  when  the  work- 
ing component  is  constant.  For  example,  suppose  that  the  body  represented 
in  Fig.  317  is  moved  along  the  line  AB  by  a  number  of  forces,  two  of  which 
(indicated)  are  constant  in  magnitude  and 

in  direction.     During   any  portion  of  the        _5__  J^''^       r--^--i 

motion,  as  from  A  to  B,  the  work  done  by      l,,,Jy,,[Zj^ 


Fi  is  Fi(AB)  and  the  work  done  by  F2  is  ' 

{F2  cos  6)  AB.  This  expression  when  written 
F  (AB  cos  6)  means  the  product  of  the  force  and  the  component  of  the  dis- 
placement along  the  line  of  action  of  the  force,  which  is  a  "  view"  of  amount 
of  work  done  by  a  force  sometimes  more  convenient  than  the  other. 

When  the  working  component  is  not  constant  in  magnitude,  then  we  may 

arrive  at  an  expression  for  the  work  somewhat  like  this:  —  Let  AB  (Fig.  318) 

be  the  path  of  the  point  of  application  of  one  of  the  forces 

acting  upon  a  body  not  shown,  and  P  any  point  on  the 

path.     Let  /'  =  the  force,  0  =  the  angle  between  F  and 

the  tangent  atP,  and  ds  =  the  elementary  portion  of  the 

path  at  P.    Then  the  work  done  by  F  during  the  elementary 

I^iG.  318  displacement  =  F  cos  4>  'ds  or  Ftds  where  Fi  means  working 

or  tangential  component  of  F;  and  the  work  done  by  F  in  the  displacement 

from  A  to  B  =  ( Ft  ds,  limits  of  integration  to  be  assigned  so  as  to  include  all 

elementary  works  Ft  ds  in  the  motion  from  A  to  B.  It  is  worth  noting  that  if 
the  force  F  acts  normally  to  the  path  at  all  points,  then  Ft  =  o  always,  and 
the  formula  gives  zero  for  the  work  done  by  F,  as  it  should. 

189 


'A 


190  Chap,  x 

The  unit  work  is  the  work  done  by  a  force  whose  working  component  equals 
unit  force  acting  through  unit  distance.*  The  unit  of  work  depends  upon  the 
units  used  for  force  and  distance;  thus  we  have  the  foot-pound,  centimeter- 
dyne,  etc.  The  second  unit  is  also  called  erg;  and  10^  ergs  =  i  joule.  The 
horse-power-hour  and  the  watt-hour  are  larger  units  of  work.  They  are  the 
amounts  of  work  done  in  one  hour  at  the  rates  of  one  horse-power  and  one 
watt,  respectively  (see  Art.  41);  thus, 

One  horse-power-hour  =  1,980,000  foot-pounds,  and 
One  watt-hour  =  3600  joules. 

When  the  works  done  by  several  forces  are  under  discussion,  it  may  be 
convenient  to  give  signs  to  their  works  according  to  this  commonly  used  rule: 
When  the  working  component  acts  in  the  direction  of  motion,  the  work  of  the 
force  is  regarded  as  positive;  when  opposite  to  the  direction  of  motion  then  the 

work  is  regarded  as  negative.     The  formula  J  F  cos  cf)  ds,  with  the  lower  and 

upper  limits  of  integration  to  correspond  to  the  initial  and  final  positions  A 
and  B,  respectively,  observes  this  rule  of  signs  for  work,  if  s  is  measured  posi- 
tive in  the  direction  of  motion  from  some  fixed  origin  to  P,  and  4>  is  measured 
from  the  "  positive  tangent"  around  to  the  line  of  action  of  the  force  as  shown 
in  the  figure.  Forces  which  do  positive  work  are  sometimes  called  efforts; 
those  which  do  negative  work  are  called  resistances. "f 

Work  Diagram.  —  If  values  of  Ft  and  5  be  plotted  on  two  rectangular  axes 
(Fig.  319)  for  all  positions  of  the  point  of  application  of  F,  then  the  curve 
joining  the  consecutive  plotted  points  might  be  called  a  "  work- 
ing force-space"  (Ft-s)   curve.      The  portion  of  the  diagram 
"under  the  curve"  (between  the  curve,  the  5-axis,  and  any 
two  ordinates)  is  called  the  work  diagram  for  the  force  F  for 
^i»a->i  "  "    '"I   *   the  displacement  corresponding  to  the   bounding  ordinates. 
„  The  area  of  a  work  diagram  represents  the  work  done  by  the 

force  during  the  displacement  corresponding  to  the  bounding 
ordinates.  Proof:  Let  m  =  the  force  scale-number,  and  n  =  the  space  scale- 
number;  that  is,  unit  ordinate  (inch)  =  m  units  of  Ft  (pounds)  and  unit 
abscissa  (inch)  =  n  units  of  5  (feet).     Also  let  A  =  area;    then 

/»ij  nhp^s        I     C^  r^   J       work 

A=    I     ydx=    I =  —  /    Ftds= 

Jx,  Ja   m  n      mnja  mn 

Hence,  A  {mn)  =  work;  that  is,  A  =  work  according  to  the  scale  number  mn 
to  be  used  for  interpreting  the  area. 

By  average  working  component  of  F  is  meant  a  value  of  Ft  which  multiplied 
by  the  distance  52  —  Si,  or  b  —  a,  gives  the  work  done  by  F.  Obviously,  that 
average  working  force  is  represented  by  the  average  ordinate  to  the  curve  of 

*  For  dimensions  of  unit  work,  see  Appendix  A. 

t  The  (negative)  work  done  by  a  resistance  on  a  body  is  often  referred  to  as  (positive) 
work  done  by  the  body  against  the  resistance. 


Art.  40 


191 


the  work  diagram.  When  that  curve  is  straight,  that  is,  when  Ft  varies  uni- 
formly with  respect  to  s,  then  the  average  working  component  equals  the  mean 
of  the  initial  and  final  values. 

Fig.  320  is  a  fac-simile  of  a  record  made  by  the  traction  dynamometer  (a 
spring  balance  essentially)  in  a  certain  train  test.  Abscissas  represent  distances 
travelled  by  the  train,  and  ordinates  represent  "  draw-bar  pulls"  (the  pulls 
between  the  tender  and  first  car  of  the  train).  Thus,  the  figure  is  a  work 
diagram.  To  determine  the  area  of  such  a  diagram  as  this  we  first  draw  in 
an  average  curve  "by  eye,"  and  then  ascertain  the  area  under  this  curve  in 
any  convenient  way. 


\//^vMv***^^'^^VV'^^^ 


•3  tons 


6  ins.  =  /  Mile. 


Fig.  320 


V     V 


Fig.  322 


§  2.  Some  Special  Cases.  —  (i)  The  work  done  by  a  force  which  is  con- 
stant in  magnitude  and  direction  equals  the  product  of  the  force  and  the  pro- 
jection of  the  displacement  of  its  point  of  application  upon  the  line  of  action 
of  the  force.  For,  let  F  =  the  force,  APB  (Fig.  321)  the  path  of  its  point  of 
application,  ^  =  the  (variable)  angle  between  ¥  and  the  direction  of  the  motion 
of  the  point  of  application  P.     Then  the  work  done  by  F  is 

I  F  cos  (j)ds  =  F  j  ds  cos  </>, 

where  ds  is  an  elementary  portion  of  the  path.     Now  ds  cos  0  is  the  projection 

of  the  element  ds  upon  F,  or  upon  any  line  parallel  to  F,  and    Cds  cos  <^  is 

the  sum  of  the  projections  of  all  the  elements  of  APB  upon  the  line.  But  the 
sum  of  the  projections  =  the  projection  of  APB  =  the  projection  of  the  chord 
AB. 

(ii)  The  work  done  by  gravity  upon  a  body  in  any  motion  equals  the  product 
of  its  weight  and  the  vertical  distance  described  by  the  center  of  gravity; 
the  work  is  positive  or  negative  according  as  the  center  of  gravity  has  descended 
or  ascended.  Let  Wi,  w^,  etc.,  denote  the  weights  of  the  particles  of  the  body; 
yii  y2 ,  etc.,  their  distances  above  some  datum  plane  —  below  which  the  body 
does  not  descend  —  at  the  beginning  of  motion;  and  ji" ,  jo",  etc.,  their  dis- 
tances above  that  plane  at  the  end  of  the  motion  (see  Fig.  322)  where  a' a"  is 
the  path  of  the  first  particle,  b'b"  that  of  the  second,  etc.).  Also  let  W  denote 
the  weight  of  the  body,  and  y'  and  y"  the  initial  and  final  heights  of  its  center 
of  gravity  above  the  plane.  Then  the  works  done  by  gravity  on  the  particles , 
respectively,  are  Wi  {yi  —  y/'),  K'2  iyi  —  y^"),  etc.,  and  the  sum  of  these 
works  can  be  written 

{wiyi  +  W2y2  +  .  .  :  )  -{wiyi"  +  W2y-i'  +...). 


192 


AP.  X 


The  first  term  of  this  sum  =  Wy',  and  the  second  =  Wy"  (see  Art.  21);  hence 
the  sum  of  these  works  done  on  all  the  particles  equals 

Wy'  -  Wy"  =  Wiy'-  y"). 
(iii)  The  algebraic  sum  of  the  works  done  by  any  number  of  forces  having  a 
common  point  of  application  during  any  displacement  of  that  point  equals  the 
work  done  by  their  resultant  during  that  displacement.  For,  let  F',  F",  F'", 
etc.,  =  the  forces,  R  =  their  resultant,  and  F/,  Ft",  F"',  etc.,  and  Rt  =  the 
components  of  the  forces  and  of  the  resultant,  respectively,  along  the  tangent 
to  the  path  of  the  point  of  application.  Now  Ft  +  F"  +  F"'  +  •  •  •  =  Rt 
(Art.  4).     Hence  Ft'  ds  +  Ft"  ds  +  F/"  ds -\-  -  •  -  =  Rtds,  and 

fFt'ds^    fFt"ds-\-    CFt'"ds-\-    •  •  •    =    fRtds; 

that  is,  the  sum  of  the  works  done  by  the  forces  equals  the  work  done  by  their 
resultant. 

(iv)   The  work  done  by  a  pair  of  equal,  colinear,  and  opposite  forces  in  any 
displacement  of  their  points  of  application  equals 


X2  r^i 

Fdr      or     -    I     F  dr 


according  as  the  forces  tend  to  separate  or  draw  the  points  of  application 

together;    F  =  the  common  magnitude  of   the  two  forces  —  not   constant 

necessarily  — ,  r  =  the  distance  between  their  points  of  appHcation,  and  n 

and  r2  =  initial  and  final  values  of  r.     Let  A  and  B  (Fig.  323)  be  the  points  of 

application  of  the  two  forces  —  acting  on  a  body 

^  V      not  shown  —  at  any  intermediate  stage  of  the 

y^- ^"_;:::^:^^^;^       displacement,  and  suppose  that  the  path  of  A  is 

■|p^^-^       /By"        ^i^^2  and  that  of  B  is  BA.    Let  x',y'  be  the 

/y'  y.      coordinates  of  A,  and  x",  y"  those  of  B.     (For 

simplicity  in  figure  we  have  taken  the  paths  of  A 
Fig  ^2^  V        J  ^ 

and  B  as  coplanar.     The  following  proof  could 

be  extended  to  cover  the  case  of  any  paths.     The  paths  are  not  necessarily 

due  to  the  forces  F  alone;  but  since  we  are  concerned  with  the  work  done  by 

these  two  forces  only,  no  mention  is  made  of  any  other  forces  concerned 

with  the  motion.)     According  to  the  preceding  paragraph  the  work  done  by 

either  force  F  in  any  displacement  equals  the  sum  of  the  works  done  by  the 

X  and  y  components  of  F  in  that  same  displacement.    Hence  in  an  elementary 

displacement  ds  the  work  of  F  on  ^  =  (-F  cos  0  dx'  -  F  sin  6  dy'),  and  the 

work  oi  F  on  B  =  {F  cos  6  dx"  -\-  FsinO  dy").      The  work  done  by  both 

forces  F  in  the  elementary  displacement  is 

F[cosd  (dx"  -  dx')  +  sin  6  {dy"  -  dy')]. 

It  will  readily  be  seen  from  the  figure  that  (x"  -  x')-  +  (y"-  y')^  =  r';  and, 
by  differentiation,  we  find  that 

(x"-  x')  {dx"-  dx')  +  {y"-  y')  {dy" -  dy')  =  rdr. 


Art.  41  195 

Dividing  by  r  and  transforming  we  find  that 

cos  6  {dx"  -  dx')  +  sin  d  {dy"  -  dy')  =  dr. 

Hence,  the  work  done  in  the  elementary  displacement  is  F  dr,  and  the  work 
done  in  the  displacement  from  AiBi  to  A1B2  equals  the  integral  of  F  dr  between 
the  limits  as  stated.  Obviously,  changing  the  senses  of  the  forces  F  (in  the 
figure)  so  that  they  tend  to  make  A  and  B  approach,  changes  the  sign  of  the 
total  work  done  by  the  forces. 

41.   Energy 

When  the  state  or  condition  of  a  body  is  such  that  it  can  do  work  against 
forces  applied  to  it,  the  body  is  said  to  possess  energy.  For  example,  a 
stretched  spring  can  do  work  against  forces  applied  to  it  if  they  are  such  that 
it  may  contract,  and  a  body  in  motion  can  do  work  against  an  applied  force 
which  tends  to  stop  it;  the  spring  and  the  body,  therefore,  possess  energy. 

The  amount  of  energy  possessed  by  a  body  at  any  instant  is  the  amount  of 
work  which  it  can  do  against  applied  forces  while  its  state  or  condition  changes 
from  that  of  the  instant  to  an  assumed  standard  state  or  condition.  The 
meaning  of  the  standard  condition  is  explained  in  subsequent  articles.  The 
unit  of  energy  must,  in  accordance  with  the  above,  be  the  same  as  the  unit  of 
work.  Thus  we  have  the  foot-pound,  foot-ton,  centimeter-dyne  (or  erg), 
the  joule,  horse-power-hour,  watt-hour,  etc.  (see  preceding  article).* 

§  I.  Mechanical  Energy.  —  Energ>^  is  classified  into  kinds  depending  on 
the  state  or  condition  of  the  body,  in  virtue  of  which  it  has  energy. 

Kinetic  Energy  of  a  body  is  energy  which  the  body  has  by  virtue  of  its 
velocity.  The  amount  of  kinetic  energy  possessed  by  a  particle  at  any  instant 
is  the  work  which  it  can  do  while  the  velocity  changes  from  its  value  at  that 
instant  to  some  other  value  taken  as  a  standard.  It  is  customary  to  take 
zero  velocity  as  the  standard  one;  this  being  understood,  then  the  amount  of 
kinetic  energy  possessed  by  a  particle  is  the  work  which  the  particle  can  do  in 
''  giving  up  all  its  velocity."  The  kinetic  energy  of  a  single  particle  whose 
mass  and  velocity  are  m  and  v,  respectively,  equals  |  mv'^. 
Proof:  Let  Fi,  Fi,  F3,  etc.,  be  the  forces  which  act  on  the 
particle  P  (Fig.  324)  and  eventually  stop  it;  and  let  AB  he 
the  path,  A  the  beginning  (where  velocity  =  Vi)  and  B  the 
end  where  velocity  =  o.  Then  we  are  to  prove  that  the 
work  done  by  the  particle  on  the  neighboring  particles  or 
bodies  (which  exert  the  forces  Fi,  F2,  F3,  etc.)  equals  |  mvi^,  during  the  motion. 
Now,  the  work  done  by  the  forces  Fi,  F2,  F3,  etc.,  on  the  particle  is 

/  Fi  cos  (f>ids  ^  I  Fa  cos  (f>2  ds  +''  '   =    /  (Fi  cos  <^i  -f-  F2  cos  02  +  •  •  •  )  ds, 
*  For  dimensions  of  a  unit  of  energy,  see  Appendix  A. 


194  Chap,  x 

where  4>i,  02,  03,  etc.,  are  the  angles  between  Fi.  F2,  F3,  etc.,  and  the  direction  of 
motion  (Art.  40).  Since  the  particle  P  exerts  forces  on  its  neighbors,  equal 
and  opposite  to  Fi,  F2,  etc.,  the  work  done  by  the  particle  on  its  neighbors  is 


/ 


(Fi  cos  01  +  F2  cos  02  +  •  •  •  )  ds. 


But  Fi  cos  01  +  F2  cos  02  +  •  •  •  =  Jnat  =  m  dv/dt,  where  at  is  the  tangential 
component  of  the  acceleration  of  the  particle;  hence  the  work  done  by  P  is 

—  /     m  (dv/dt)  ds  ^  —   j    m  {ds/dt)  dv  =  —    j    mv  dv  =  ^  mv^. 

The  kinetic  energy  of  a  body  (a  collection  or  system  or  particles)  is  the 
sum  of  the  kinetic  energies  of  the  constituent  particles  of  the  body.  We  will 
now  evaluate  this  sum  for  certain  common  cases,  —  namely,  (i)  translation, 
(ii)  rotation,  and  (iii)  combined  translation  and  rotation. 

(i)  In  translatory  motion  all  particles  of  the  moving  body  have  at  each 
instant  equal  velocities;  hence,  the  simi  of  the  kinetic  energies  of  the  particles 
is  \  ntiv-  +  I  WoZ'-  +  •  •  •  =  2  "^^  (2^0,  where  Wi,  nh,  etc.,  =  the  masses  of 
the  particles  and  v  =  their  common  velocity  at  the  instant  imder  consideration. 
Or,  if  M  =  the  mass  of  the  body  and  E  =  energy,  then 

E  =  ^Mv^  =  i(W/g)v\  (i) 

If  32.2  is  written  for  g,  then  v  should  be  expressed  in  feet  per  second.  E  will 
be  in  foot-pounds,  foot-tons,  etc.,  according  as  W  is  expressed  in  pounds,  or 
tons,  etc. 

(ii)  In  a  rotation  about  a  fixed  axis  the  velocity  of  any  particle  of  the  body 
equals  the  product  of  the  angular  velocity  of  the  body,  expressed  in  radians  per 
unit  time,  and  the  distance  from  the  particle  to  the  axis  of  rotation  (Art.  37). 
Hence,  the  sum  of  the  kinetic  energies  of  the  particles  of  the  body  is 

I  mi  (ri  co)2  -f  i  1712  (r^f^y  +  •  •  •   =  |  oj^  Zmr-, 

where  co  =  the  angular  velocity  of  the  body  at  the  instant  under  consideration, 
and  ri,  ^2,  etc.,=  the  distances  of  the  particles  respectively  from  the  axis  of 
rotation.  But  I,mr^  =  the  moment  of  inertia  of  the  body  about  the  axis  of 
rotation;   hence,  the  kinetic  energy  is  given  by 

£  =  i  /co2  =  i  Mk^o:'  =  h  (W/g)  kW',  (2) 

where  I  =  the  moment  of  inertia  described,  and  k  =  the  radius  of  gyration  of 
the  body  about  the  axis  of  rotation.  If  32.2  is  written  for  g,  then  k  should  be 
expressed  in  feet  and  co  in  radians  per  second  (w  —  2  -wn  where  n  =  revolu- 
tions per  second).  Then  E  will  be  in  foot-pounds,  foot-tons,  etc.,  according 
as  W  is  expressed  in  pounds,  tons,  etc. 

(iii)  A  body  which  has  a  combined  plane  translation  and  rotation  (Art.  50), 
Uke  a  wheel  rolling,  has  kinetic  energy  given  by 

£  =  i  M^^  +  Hco^  =  I  {W/g)  v'  +  i  (W/g)  ¥oi\  (3) 


Art.  41  195 

where  M  =  mass  of  the  body,  v  =  velocity  of  the  center  of  gravity,  /  =  the 
moment  of  inertia  of  the  body  with  respect  to  an  axis  through  the  center  of 
gravity  perpendicular  to  the  plane  of  the  motion,  k  =  radius  of  gyration  with 
respect  to  the  same  axis,  and  w  =  the  angular  velocity  of  the  motion.  Proof 
of  this  formula  is  given  in  Art.  51.  The  portions  ^  Mv^  and  |  /co^  of  the  kinetic 
energy  are  sometimes  called  the  translational  and  rotational  components, 
respectively. 

As  an  example  of  the  use  of  the  preceding  formula  we  find  the  kinetic  energy 
of  a  cylindrical  disk,  6  feet  in  diameter  and  400  pounds  in  weight,  which  is 
rolling  so  that  the  center  has  a  velocity  of  4  feet  per  second.  M  =  400  -i-  32.2 
=  12.4  slugs;  the  square  of  the  radius  of  gyration  of  the  disk  is  |  3^  =  4.5  feet^ 
(see  Art.  36);  and  cxi,  the  rate  at  which  the  wheel  is  turning,  is  4  -^3  radians  per 
second.     Hence 

£  =  -  12.4  X  4"  +  -  12.4  X  4.5(4  -^  3)'"  =  148.8  foot-pounds. 

Potential  Energy.  —  A  body  may  possess  energy  which  is  not  due  to  velocity. 
Thus  two  mutually  attracting  bodies  can  do  work  against  forces  applied  to 
either  or  both  if  allowed  to  move  so  that  they  approach  each  other;  and,  as 
stated,  a  compressed  or  stretched  spring  can  do  work  against  applied  forces 
if  permitted  to  resmne  its  natural  length.  The  "change  of  condition  or  state" 
in  the  first  case  is  a  change  in  configuration,  a  change  in  the  positions  of  the 
bodies  relative  to  each  other;  and,  in  the  second  case,  if  we  conceive  of  the 
spring  as  consisting  of  discrete  particles,  the  change  is  also  one  in  configuration 
(of  the  particles).  Energy  of  a  system  of  particles  dependent  on  configura- 
tion of  the  system  is  called  energy  of  configuration,  and  potential  energy 
more  commonly. 

The  amount  of  potential  energy  possessed  by  a  system  in  any  configuration 
is  the  work  which  it  can  do  in  passing  from  that  configuration  to  any  other 
taken  as  a  standard,  it  being  understood  that  no  other  change  of  condition 
takes  place.  The  standard  configuration  may  be  chosen  at  pleasure,  but  it  is 
convenient  to  so  select  it  that  in  all  other  configurations  considered  the  poten- 
tial energy  is  positive. 

A  most  common  case  of  potential  energy  is  that  of  the  earth  and  an  elevated 
body.  In  this  case,  standard  configuration  means  one  in  which  the  body  and 
earth  are  as  near  together  as  possible.  Practically,  it  is  necessary  to  regard 
the  earth  as  fixed  and  the  energy  as  resident  in  the  elevated  body.  The  amount 
of  potential  energy  of  an  elevated  body  is  just  equal  to  the  work  which  gravity 
would  do  upon  the  body  during  the  descent  into  the  standard  or  lowest  position, 
and  this  work  is  given  by  Wh  (see  preceding  article),  where  W  =  the  weight 
of  the  body  and  h  =  the  distance  through  wliich  the  center  of  gravity  of  the 
body  can  descend. 

§  2.  Other  Forms  of  Energy.  —  Kinetic  energy  and  potential  energy 
are  often  called  mechanical  energy.     It  is  the  opinion  of  some  that  all  energy 


Iq6  Chap,  x 

is  mechanical,  and  some  think  that  it  is  all  kinetic.  Whether  either  of  these 
views  be  correct,  it  is  practically  necessary  to  recognize  other  forms.  A  mere 
enumeration  of  these  with  brief  remarks  is  sufficient  for  the  present  purpose, 
since  we  shall  deal  mostly  with  energy  known  to  be  mechanical. 

Thermal  Energy.  —  A  hot  body  may  do  work  under  favorable  conditions; 
thus,  if  such  a  one  is  placed  in  a  boiler  containing  water,  the  water  will  be 
heated  and  a  part  may  be  converted  into  steam  which  may  drive  a  steam 
engine,  that  is,  do  work.  By  giving  up  its  heat  the  hot  body  has  done  work, 
and,  hence,  by  definition,  it  possessed  energy  in  its  heated  state.  Not  only  is 
this  fact  well  known,  but  also  the  fact  that  a  given  quantity  of  heat  represents 
a  definite  amount  of  energy.  Thus,  one  British  thermal  unit  (B.T.U.),  which 
is  the  amount  of  heat  required  to  raise  the  temperature  of  one  pound  of  water 
one  Fahrenheit  degree,  =  778  foot-pounds.  And  one  (small)  calorie,  which 
is  the  amount  of  heat  required  to  raise  the  temperature  of  one  gram  of  water 
one  Centigrade  degree,  =  4.187  X  10^  ergs  (at  15  degrees).  Based  on  the 
molecular  hy^Dothesis  the  common  theory  is  that  heat  is  due  to  the  vibratory 
motion  of  molecules,  that  is,  thermal  energy  is  kinetic. 

Chemical  Energy.  —  Many  substances  combine  chemically,  and  their  com- 
bination gives  evidence  that  they  possessed  energy.  Thus,  coal  and  oxygen 
combine  and  produce  heat  which,  as  we  have  seen,  is  a  form  of  energy.  We 
rightly  say,  therefore,  that  the  coal  and  oxygen  before  combination  possessed 
energy.  Based  on  this  molecular  hyj^othesis  the  theory  of  chemical  energy 
in  cases  where  heat  is  generated  in  the  chemical  combination  is  that  internal 
(molecular)  forces  of  the  substances  do  work  during  the  combination,  and, 
hence  (see  Art.  43),  increase  the  kinetic  energy  of  the  molecules.  According 
to  this  explanation  the  energy  before  combination  is  potential;  and  after, 
kinetic. 

Electrical  Energy.  —  If  a  charged  storage  battery  be  connected  with  a  motor, 
work  may  be  done  by  the  latter.  As  the  work  is  done,  the  electrical  condition 
of  the  battery  changes,  and  we  therefore  ascribe  the  energy  to  the  batter}'. 
The  energy  is  called  electrical  because  it  is  due  to  a  change  of  electrical  condi- 
tion. The  nature  of  electrical  energy  is  even  less  understood  than  that  of 
thermal  energy,  and  no  commonly  accepted  explanation  of  it  has  yet  been  made^ 

42.   Power 

§  I .  In  common  parlance  the  word  power  has  many  meanings  (see  diction- 
ary). Thus  we  hear  of  the  power  of  a  giant,  power  of  example,  power  of  the 
press,  etc.  And  of  things  mechanical,  we  hear  such  expressions  as  a  powerful 
derrick,  a  powerful  cannon,  a  powerful  pump,  etc.  On  reflection  we  note  that 
the  adjective  in  these  three  expressions  probably  does  not  refer  to  the  same 
feature  of  the  derrick,  cannon,  and  pump.  A  derrick  is  probably  called  power- 
ful because  it  can  lift  a  very  heavy  body,  or  exert  a  very  great  (lifting)  force. 
A  cannon  is  generally  called  powerful  because  it  can  project  a  heavy  shot 


Art.  42  197 

with  great  velocity,  and  we  shall  see  presently  such  performance  depenas  on 
the  energy  which  the  gun  can  impart  to  the  shot.  A  pump  is  probably  called 
powerful  because  it  can  elevate  or  transport  a  large  quantity  of  liquid  in  a  short 
space  of  time,  or  perform  much  work  per  unit  time. 

Use  of  the  word  power  in  the  sense  of  force  was  very  common  in  engineering 
literature  at  one  time.  Such  usage  is  comparatively  rare  now,  but  not  obso- 
lete. Thus  we  read  of  the  "  tractive  power  of  a  locomotive"  to  denote  pull  in 
the  bulletins  of  the  American  Locomotive  Company.  (But  Goss  in  his  Loco- 
motive Performance,  and  Henderson  in  his  Locomotive  Operation,  seem  to  prefer 
tractive  force ;  and  in  Locomotive  Tests  and  Exhibits,  of  the  Pennsylvania  Rail- 
road System  at  the  Louisiana  Purchase  Exposition,  we  find  "  tractive  effort " 
to  denote  that  pull.)  The  other  two  uses  of  the  word  power  to  denote  (i) 
work  or  energy,  and  (ii)  rate  at  which  work  is  done  or  energy  is  transmitted 
or  transformed  are  c[uite  common.  Thus  in  the  same  text-book  we  find: 
(i)  "  the  actual  power  utilized  is  one-half  the  energy  available,"  and  (ii)  "the 
power  of  the  plant  is  about  470  horse-power"  (258,500  foot-pounds  per 
second,  see  below).  And  in  another  book  there  appear:  (i)  "  the  power  of 
the  rotating  shaft  could  be  converted  into  electrical  energy,"  and  (ii)  '"the 
power  is  here  measured  in  kilowatts  "  (one  kilowatt  equals  10'"  ergs  per  second, 
see  below).  It  seems  probable  that  this  double  usage  of  the  word  power  in 
engineering  literature  will  persist.  In  common  with  most  authors,  even  those 
quoted  above,  we  will  define  power  in  a  single  sense,  namely,  —  as  the  rate  at 
which  work  is  done. 

Units  of  Power  *  like  units  of  work  may  be  classed  into  gravitational,  which 
vary  slightly  with  locality,  and  absolute.  Thus,  the  foot-pound  per  minute  and 
the  kilogram-meter  per  second  are  units  of  the  first  class ;  also  the  (practical) 

EngUsh  and  American  horse-power  =  550  foot-pounds  per  second 

=  33,000  foot-pounds  per  minute, 

Continental  horse-power  ==^75  kilogram-meters  per  second 

=  4500  kilogram-meters  per  minute. 

The  dyne-centimeter  (or  erg)  per  second  is  a  unit  of  the  second  class;  also  the 
watt  which  is  10'^  ergs  per  second,  and  the  (practical) 

kilowatt  —  1000  watts  =  io^°  ergs  per  second. 

The  Bureau  of  Standards  has  recently  decided  to  adopt  the  English  and 
American  horse-power  as  the  exact  equivalent  of  746  watts,  thus  making 
this  horse-power  an  absolute  unit.  "  Thus  defined  it  is  the  rate  of  work  ex- 
pressed by  550  foot-pounds  per  second  at  50°  latitude  and  sea  level,  approxi- 
mately the  location  of  London,  where  the  original  experiments  were  made  by 
James  Watt  to  determine  its  value.  The  '  continental  horsepower '  is  similarly 
most  conveniently  defined  as  736  watts,  equivalent  to  75  kilogram-meters  per 
second  at  latitude  52°  30',  or  Berlin."  f 

*  For  dimensions  of  a  unit  of  power  see  Appendix  A. 
t  Circular  of  the  Bureau  of  Standards,  No.  34. 


198  c^^-  '^ 

§  2.  Measurement  of  Power.  —  There  is  only  one  instrument  in  common 
use  which  measures  power  directly,  the  wattmeter.  It  measures  electric 
power  and  reads  in  watts,  hence  the  name  wattmeter.  Power  other  than  elec- 
trical is  generally  measured  indirectly  by  measuring  the  amount  of  work  done 
or  energy  transmitted  in  a  certain  length  of  time;  this  work  or  energy  divided 
by  the  time  gives  the  average  power  for  the  period.  And  to  measure  the 
work  or  energy  generally  requires  the  measurement  of  a  force;  this  force  multi- 
pUed  by  the  distance  through  which  it  acts  (as  explained  later)  gives  the  work 
or  energy.  Thus  most  appliances  for  ascertaining  power  measure  force  first 
of  all,  and  so  are  properly  called  dynamometers  (force-measurers).  Dynamom- 
eters are  of  two  kinds,  —  absorption  and  transmission.  Those  of  the  first 
kind  absorb  or  waste  the  energy  which  they  measure,  and  those  of  the  second 
kind  transmit  the  energy  or  nearly  all  of  it.  A  great  many  dynamometers 
have  been  devised.     Only  one  of  each  kind  is  here  described.* 

Prony  Brake.  —  k  simple  form  is  shown  in  Fig.  325.  AA  are  two  bearing 
blocks  which  bear  against  the  face  of  the  pulley  on  the  shaft  of  the  motor  or 
other  machine  whose  power  is  to  be  measured;  BC  is  the  beam,  one  end  of 
which  is  supported  on  a  post  D  which  rests  on  the  platform  of  a  weighing  scale; 
BB  are  nuts  by  means  of  which  the  pressures  between  the  pulley  and  the  bear- 
ing blocks  may  be  changed  and  consequently  the  frictional  drag  also  when 
the  pulley  is  turning.  The  drag  on  the  brake  tends  to  depress  the  end  C 
when  the  pulley  is  rotating  as  indicated. 


'w/!nu//f»'- 


Fig.  325 


Fig.  326 


Let  5  =  the  reading  of  the  scale  when  the  pulley  is  rotating  at  the  desired 
speed,  the  brake  then  absorbing  the  energy  which  is  to  be  measured;  n  =  the 
revolutions  of  pulley  per  unit  time;  a  =  the  horizontal  distance  from  the 
support  of  C  to  the  center  of  the  shaft;  and  X  =  a  correction  explained  below. 
Then  the  power  equals 

P  =  (5  -  X)  2  -Kan.  (i) 

If  S  and  X  are  expressed  in  pounds,  a  in  feet,  and  n  in  revolutions  per  minute, 

then 

P  =  0.000190  (5  —  X)  an    horse-power.  (2) 

The  meaning  of  X  will  appear  from  the  following  derivation  of  formula  (i). 
Let  F  =  the  total  frictional  drag  on  the  pulley  while  the  energy  to  be  measured 

*  For  full  descriptions  of  many  others  see  Flathers'  Dynamometers  or  Carpenter  and  Deder- 
ichs'  Experimental  Engineering. 


Art.  42 


199 


is  being  absorbed  and  d  =  diameter  of  pulley.  The  work  done  on  the  pulley 
by  this  friction  per  revolution  is  F  ird,  and  per  unit  time  the  work  is  Fir  dn. 
Now  let  W  =  weight  of  the  brake,  and  w  =  weight  of  D;  then  it  is  plain  from 
Fig.  326  that  \Fd  +  Wb  =  (S  -  w)  a,  or 

Fd  =  2[S  -  {w  +  Wb/a)]  a;  and  hence  P  =  [S  -  (w -\-  Wb/a)]  2  wan. 

This  last  equation  is  like  (i)  except  that  X  replaces  w  +  Wb/a.  Now  obvi- 
ously Wb/a  is  the  pressure  on  the  scale  due  to  W;  hence  X  is  that  portion  of 
S  due  to  W  and  w.  X  can  be  determined  directly  as  follows:  Loosen  the 
screws  BB  and  insert  a  small  roller  between  the  top  of  pulley  and  the  upper 
block  A,  but  without  shifting  C;  then  read  the  scale.  That  reading  =  X, 
for  the  pressure  on  the  scale  then  =  w  +  Wb/a. 


H.-j^f^^-y  --V, 


t....^x....J 


a 


k-b-H      V-'b  -H 


Fig.  327 


Fig.  328 


Tatham  Dynamometer.  —  This  consists  of  four  pulleys,  A,  B,  C  and  D 
(Fig.  327),  two  levers  E  and  F,  a  weighing  beam  G,  and  a  belt  HIJK.  Pulleys 
A  and  B  are  mounted  on  the  frame  of  the  dynamometer;  pulleys  C  and  D  are 
idlers  and  are  mounted  on  the  levers  which,  in  turn,  are  supported  on  knife 
edges  resting  on  the  frame  and  by  knife-edge  Hnks  L  and  M  suspended  from 
the  weighing  beam,  all  as  shown;  the  weighing  beam  is  supported  from  the 
frame  at  N.  The  dimensions  are  such  that  the  straight  portions  of  the  belt 
are  vertical,  and  H  and  K  are  vertically  below  the  knife-edge  supports  of  the 
levers.  The  shafts  of  A  and  B  extend  backwards  to  connect  with  machines 
between  which  the  energy  to  be  measured  is  transmitted.  In  all  cases,  the 
connections  to  machines  should  be  made  so  that  the  tension  in  I  is  greater 
than  that  in  /(/"tight"  and  /  "slack");  and,  if  possible,  the  machine 
whose  power  is  to  be  determined  should  be  connected  to  or  be  on  the  shaft  of 
A .  When  the  dynamometer  is  in  operation,  then  L  and  M  pull  on  the  weighing 
beam;  and,  if  the  beam  be  balanced  by  the  poise,  then  the  scale-reading  gives 
the  difference  in  tensions  of  /  and  /,  or  P2  -  P\  (see  Fig.  328).    Let  5  =  the 


200  Chap,  x 

scale-reading,  n  =  revolutions  per  unit  time  of  pulley  A,  D  ^  diameter  of 
the  pulley  plus  thickness  of  belt,  and  P  =  power;  then 

P  =  STrDn. 

For  Sir  D  is  the  work  done  by  the  belt  on  A  in  one  turn  of  A ;  and,  hence,  the 
work  done  per  unit  time  is  Sir  Dn. 

Fig.  328  shows  the  forces  acting  on  the  various  parts,  and  makes  plain  how 
the  poise  measures  P2  —  Pi-  Thus,  from  the  right-hand  lever  Qi  =  Pic/b; 
from  the  left-hand  lever  Q2  =  P2c/b;  and  from  the  weighing  beam  Wx  = 
(Q2  -Qi)a=  (P2  -  Pi)  ac/b.  Hence,  P2  -  Pi  =  {Wb/ac)  x.  Now,  Wb/ac 
is  a  constant,  and  so  it  is  possible  to  graduate  the  scale  beam  (mark  values  of 
X  on  it),  so  that  the  readings  will  give  the  corresponding  values  of  P2  —  Pi. 
(No  mention  has  been  made  of  the  weights  of  the  parts.  These  are  counter- 
balanced by  a  balancing  weight  on  the  scale  beam  as  in  an  ordinary  platform 
scale.) 

§  3.  Indicator;  Locomotive  Power.  —  To  determine  the  work  done  in 
the  cylinder  of  a  steam  or  gas  engine  per  stroke  or  per  unit  time,  use  is  made 
of  an  instrument  called  an  indicator.  The  indicator  makes  a  diagram  or 
"  card"  from  which  the  intensity  of  the  pressure  on  either  side  of  the  piston 
at  any  point  of  a  stroke  can  be  read.  Fig.  329  represents,  in  principle,  the 
original  form  of  indicator  as  used  by  James  Watt  (1736-1819). 
A  is  a  cylinder;  5  is  a  piston  working  against  a  coil  spring  C 
whose  upper  end  is  fixed;  D  is  a,  pencil  which  presses  against 
the  card  or  paper  E;  F  is  a  frame,  movable  right  and  left  in 
suitable  sHdes,  for  holding  the  paper  or  card.  When  the  piston 
is  moved  the  pencil  simply  makes  a  vertical  line  on  the  card; 
when  the  frame  is  moved  the  pencil  makes  a  horizontal  line. 
To  take  a  diagram  the  cylinder  of  the  instrument  is  connected 
with  one  end  of  the  cylinder  of  the  engine  to  be  indicated, 
and  the  frame  is  connected  to  the  cross-head  of  the  engine  with 
suitable  reducing  device  so  that  the  frame  gets  a  motion  just  like 
that  of  the  piston  but  greatly  reduced.  When  the  instrument 
is  connected  up,  as  just  described,  then  the  pencil  describes  a  curve,  something 
like  GHIJG,  the  upper  portion  GHI  being  drawn  during  the  forward  stroke 
and  the  lower  portion  IJG  during  the  return.  The  ordinates  to  the  curve 
from  the  line  of  zero  pressure  K  represent  pressure  per  unit  area  in  the 
cylinder,  the  scale  of  ordinates  depending  on  the  stiffness  of  the  spring  of 
course.  The  horizontal  width  of  the  diagram  represents  the  stroke  of  the 
piston. 

Fig.  330  is  a  facsimile  of  an  indicator  card;  the  solid  curve  pertains  to  one 
end  of  the  cylinder,  and  the  dotted  curve  to  the  other  end;  AB  is  the  line  of  zero 
pressure.  The  area  AC  DEB  A  represents  the  work  done  on  one  side  of  the 
piston  (per  unit  area)  during  the  forward  stroke,  and  the  area  BEFCAB 
represents  the  work  done  on  it  during  the  return  stroke.     But  the  first  work  is 


Art.  42  2°^ 

positive,  and  the  second  negative;  hence  the  work  done  on  that  side  of  the 
piston  during  both  strokes  is  represented  by  the  area  enclosed  by  the  curve 
CDEFC.  Similarly,  the  area  of  the  dotted 
curve  represents  the  work  done  upon  the 
other  side  of  the  piston  (per  unit  area) 
during  a  to-and-fro  stroke.  The  mean 
heights  of  these  areas  represent  pressures 
per  unit  area  which  are  called  mean  effective  ^iq.  330 

pressures,  one  for  the  head-end  and  one  for 

the  crank-end  of  the  cylinder.  Let  pi  =  mean  effective  pressure  for  the  head- 
end, p2  =  that  for  the  crank-end,  A  =  area  of  cross-section  of  the  cylinder, 
A'  =  area  of  cross-section  of  the  piston  rod,  and  I  =  length  of  stroke.  Then 
the  work  done  by  the  steam  in  the  head-end  during  two  consecutive  strokes 
=  piAl;  that  done  by  the  steam  in  the  crank-end  =  p2  {A  —  A')  I,  and  the 
total  work  done  is  the  sum  of  these  expressions. 

The  average  of  the  mean  effective  pressures  {pi  and  P2)  for  the  two  ends  of 
the  cylinder  is  sometimes  called  the  mean  effective  pressure  (for  the  cylinder). 
Let  p  =  this  mean  effective  pressure  (per  unit  area) ;  a  =  the  average  of  the 
areas  of  the  two  sides  of  the  piston,  or  what  amounts  to  the  same  thing,  the 
area  of  the  cross-section  of  the  cylinder  minus  one-half  the  area  of  the  cross- 
section  of  the  piston  rod;  n  =  the  number  of  strokes  of  the  piston  per  unit 
time;  and  /  =  length  of  stroke,  as  before.  Then,  as  will  be  shown  presently, 
the  work  done  on  the  piston  per  double  stroke  is  2  pal  closely;  and,  hence,  the 
work  done  per  unit  time,  or  the  power,  is 

F  =  plan.  (i) 

If  the  customary  units  are  used,  namely,  p  in  pounds  per  square  inch,  /  in  feet, 
a  in  square  inches,  and  n  in  strokes  per  minute,  then  P,  above,  is  in  foot-pounds 

per  minute;  and  a^^^ 

P  =  — horse-powers. 

33,000 

To  justify  2  pal:  —As  already  explained,  the  work  done  in  the  cylinder  per 
double  stroke  equals  p^^i  _[_  p^  (^1  -  A')  I. 

This  can  be  written  as  follows: 

[{Pi  +  P2)A-P2A']1,    or     ^  X  l(Pi -^  P2)  (a  -  j-^^A'Y 

Now  pi  =  p2  nearly,  and  therefore  p2  ^  (pi  +  P2)  =  ^  nearly.     Hence  the 
work  done  per  double  stroke  equals  approximately 

2  X  I  {pi-h  P2)  {A  -h  ^')  I  or  2  pal. 
For  a  single-expansion,  two-cylinder  locomotive,  P  =  2  plan.     Let  5  =  the 
"  piston  speed,"  the  actual  distance  which  a  piston  describes  in  its  cylinder 
per  unit  time;  then  s  =  In  and 

P  =  2  pas.  (2) 


202 


Chap,  x 


With  customary  units  for  p,  a,  and  5  (pounds  per  square  inch,  square  inches, 
and  feet  per  minute  respectively)  the  foregoing  formula  gives  P  in  foot-pounds 
per  minute.  Since  the  piston  speed  and  the  velocity  of  the  locomotive  are 
related,  it  is  possible  to  express  the  indicated  power  of  a  locomotive  in  terms 
of  its  velocity.  Thus  let  v  =  the  velocity  of  the  locomotive,  and  D=  diam- 
eter of  the  driving  wheels;  then  one  turn  of  the  drivers  means  a  displacement 
of  the  locomotive  equal  to  irD  and  a  displacement  of  the  piston  relative  to 
its  cylinder  equal  to  2  /.  Hence  v/s  =  xD/2  /,  or  5  =  (2  l/irD)  v.  Substitut- 
ing for  s  in  the  preceding  formula  for  P,  we  find  that 


4  pal  I 


(3) 


100 


80 


•£60 
-o 
« 

;q-4o 


where  c?  =  diameter  of  the  cylinder.  (Strictly  d  =  the  diameter  of  a  circle 
whose  area  equals  the  area  of  the  cross-section  of  the  cyUnder  minus  one-half 
the  area  of  the  cross-section  of  the  piston  rod.)  With  pounds  per  square  inch 
for  p,  inches  for  d,  I,  and  D,  and  feet  per  minute  for  v,  the  foregoing  formula 
gives  P  in  foot-pounds  per  minute.  Both  formulas  for  P  show  that  the  power 
of  a  locomotive  is  zero  at  starting,  and  would  increase  exactly  with  the  velocity 
if  the  mean  effective  pressure  were  the  same  at  all  speeds. 

The  mean  effective  pressure  depends  upon  the  boiler  pressure  obviously, 
and  on  the  cut-olT  and  piston  speed.*     The  American  Locomotive  Company 

has  adopted  the  line  A  BCD  (Fig.  331),  as 
expressing  the  variation  of  mean  effective 
pressure  with  change  of  piston  speed,  for 
the  manner  of  running  (cut-ofif,  etc.)  which 
engine  men  usually  employ.  Thus,  for  all 
speeds  up  to  250  feet  per  minute,  the  mean 
effective  pressure  is  taken  at  85  per  cent  of 
the  boiler  pressure;  at  500  feet  per  minute, 
it  is  taken  at  about  65  per  cent,  etc.  Let 
po  =  boiler  pressure  and  K  =  ratio  of  mean 
effective  to  boiler  pressure,  which  may  be 
called  speed  coefficient  for  convenience;  so 
that  p  =  Kpo.  Then  the  formula  for  indicated  power  of  the  locomotive  can 
be  written  p  _  2  A^o^^.  (4) 

Thus,  for  a  given  boiler  pressure  the  power  varies  as  Ks.  The  line  OEFGH 
(Fig.  331)  is  a  graph  of  the  preceding  equation,  the  maximum  value  of  P  being 
called  100  per  cent.  It  appears,  then,  that  for  the  American  Locomotive 
Company  speed  coefficients,  the  power  increases  uniformly  up  to  a  piston  speed 
of  250  feet  per  minute,  then  less  rapidly  up  to  a  maximum  value  at  about 
700  feet  per  minute,  then  remains  nearly  constant  up  to  about  1000  feet  per 
minute,  and  then  diminishes. 

*  See  Fig.  42  in  Goss'  High  Pressures  in  Locomotive  Service,  which  shows  clearly  how  the 
mean  effective  pressure  varied  in  a  test  made  by  him. 


?0 


X 

'f 

~G* 

■~lr 

A^S 

-.B 

/ 

\X 

< 

/e 

\ 

— ^ 

N9 

1 

>* 

>s. 

/ 

^^ 

"v^^ 

/ 

^s 

^D 

/ 

^ 

I 

0 

600 
Feet     per 


1?00 
Minute. 


Art.  43  203 

43.   Principles  of  Work  and  Energy 

§  I.  Principle  of  Work  and  Kinetic  Energy.  —  In  any  displacement 
of  a  single  particle  the  forces  acting  upon  it,  if  any,  do  more  or  less  work;  and, 
in  general,  the  velocity  of  the  particle  is  changed,  and,  hence,  the  kinetic  energy 
also.  There  is  a  simple  relation  between  the  total  work  done  upon 
the  particle  by  all  the  forces  acting  upon  it  in  the  displacement 
and  the  change  in  the  kinetic  energy  as  we  will  now  show.  Let 
P  (Fig.  332)  be  the  particle;  m  =  its  mass;  OAB  be  its  path  (not 
a  plane  curve  necessarily);  Vi  =  its  velocity  at  A,  and  v^  =  its 
velocity  at  B;  R  =  the  resultant  of  all  the  forces  acting  on  P; 
and  Rt  =  the  component  of  R  along  the  tangent  to  the  path  at  P. 
Then  the  work  done  by  all  the  forces  during  an  elementary  dis- 
placement ds  is  Rt  ds.  But  Rt  =  mat  =  ^  dv/dt,  where  a  =  tan- 
gential component  of  the  acceleration  of  P.  Hence  the  work  done  on  P  in 
the  displacement  ds  is  m  (dv/dt)  ds  =  m  (ds/dt)  dv  =  mv  dv;  and  the  work 
done  in  the  total  displacement  AB  is 

mvdv  =  ^  mvi^  —  \  mv^. 

Now  ^  mvz^  is  the  kinetic  energy  of  the  particle  at  B,  and  |  mvi^  is  its  kinetic 
energy  at  A ;  hence  |  mvz^  —  §  nivi^  is  the  increment  in  the  kinetic  energy  of 
P.  Thus  we  have  the  simple  relation,  —  in  any  displacement  of  a  particle, 
the  work  done  by  all  the  forces  acting  upon  it  equals  the  increment  in  the 
kinetic  energy  of  the  particle.  If  the  total  work  done  is  positive  then  the 
increment  in  the  kinetic  energy  is  positive  also,  and  there  is  a  real  gain  and 
increase  in  velocity;  if  the  total  work  done  upon  the  particle  is  negative,  then 
the  increment  in  the  kinetic  energy  is  negative  and  there  is  a  loss  and  decrease 
in  velocity. 

Let  Pi,  P2,  P3,  etc.,  be  the  particles  of  any  body  (not  rigid  necessarily).  In 
any  displacement  of  the  body, 

work  done  by  forces  acting  upon  Pi  —  increment  in  kinetic  energy  of  Pi, 

(<  a         <(  i(  li  "        p     __  <'  '<  <<  "  'I   p 

II  a        ((  a  (I  a       jj  c(  a         u  a  ic    n 

-f  3   —  -13, 

etc.         =  etc. 

Adding  we  get  total  work  done  on  all  particles  =  sum  of  increments  in  their 
kinetic  energies  =  increment  in  kinetic  energy  of  the  system.  That  is,  in  any 
displacement  of  any  body  the  total  work  done  upon  it  by  all  the  external  and  internal 
forces  acting  upon  it  equals  the  increment  in  the  kinetic  energy  of  the  body. 

In  a  displacement  of  a  rigid  body  the  total  work  done  by  the  internal  forces 
equals  zero.  Proof:  —  Consider  any  internal  force  exerted,  say,  on  Pi  by  P2; 
Pi  exerts  an  equal,  opposite,  and  colinear  force  on  Po.  Since  the  body  is  rigid 
the  distance  between  the  points  of  application  {Pi  and  P2)  of  these  two  forces 


204  Chap,  x 

does  not  change,  and  hence  (Art.  40)  the  total  work  done  by  these  two  forces 
equals  zero.  But  all  the  internal  forces  occur  in  such  pairs;  hence,  the  total 
work  done  by  all  the  internal  forces  equals  zero,  as  stated.  Thus  we  have 
the  principle,  —  in  any  displacement  of  a  rigid  body  the  total  work  done  upon 
the  body  by  the  external  forces  acting  upon  it  equals  the  increment  in  the  kinetic 
energy  of  the  body. 

From  these  principles  it  follows  that  the  rate  at  which  work  is  done  upon  a 
body  equals  the  rate  at  which  it  gains  kinetic  energy.  But  the  rate  at  which 
work  is  done  is  power;  so  we  may  state  that  the  combined  power  of  all  the  forces 
doing  work  on  a  body  at  any  instant  equals  the  rate  at  which  it  is  gaining  kinetic 
energy  then. 

The  foregoing  principles  written  out  mathematically  would  take  the  form: 
work  done  =  increment  in  kinetic  energy.  Since  work  is  of  the  form  force 
X  distance  or  space,  we  may  state  that  the  "  space-efifect "  of  force  is  kinetic 
energy.  (The  "  time-effect"  of  force  is  momentum,  see  Art.  45.)  The  fore- 
going principles  are  especially  well  adapted  for  ascertaining  the  change  in 
velocity  —  velocity-square,  rather  —  when  it  is  possible  to  compute  the  total 
work  done  on  the  body  under  consideration  for  the  space  in  which  the  change 
takes  place.  By  their  means  we  may  ascertain  also  something  about  the  forces 
or  displacement  which  accompany  any  given  change  in  the  kinetic  energy  of 
a  body.     We  illustrate  by  means  of  some 

Examples.  —  i.  .4  (Fig.  2>2>z)  is  a  body  weighing  400  pounds.  It  is  dragged 
along  a  rough  horizontal  plane  B  by  a  force  P,  inclined  as  shown;  P  =  80 
pounds.  The  coefficient  of  friction  is  about  i/io.  What  is  t)ie 
^  velocity  acquired  from  rest  in  the  first  10  feet?  In  the  first 
^'--      20  feet?     The  normal  pressure  between  A  and  B  =  400  —  80 


•"f/iii/inii' 
B 


Fig.  333 


sin  20°  =  372.6  pounds;  hence,  the  friction  =  37.3  pounds.  Now' 
we  know  all  the  forces  acting  on  A .  Gravity  (400  pounds)  does 
no  work  on  A ;  the  work  done  by  P  during  a  displacement  of 
ID  feet  =  (80  cos  20°)  X  10  =  752  foot-pounds;  the  reaction  of  B  on  A  does 
work  =  —  37.3  X  10  =  —  373  foot-pounds.  Hence,  the  total  work  done  on 
yl  =  y52  —  373  =  379  foot-pounds;  and  this  is  also  the  amount  of  the  gain 
in  the  kinetic  energy  of  A  during  10  feet  of  displacement.  Let  Vi  =  the 
velocity  (in  feet  per  second)  at  the  end  of  the  first  10  feet;  then  the  kinetic 
energy  of  A  at  the  end  of  the  first  10  feet  =  |  (400/32.2)  ^r  =  6.21  Vi^  foot- 
pounds. Hence  6.21  Vi^  =  379,  or  Vi  =  7.81  feet  per  second.  Let  V2  =  the 
velocity  of  A  at  the  end  of  the  first  20  feet;  then  the  energy  of  A  there  =  6.21 
V2^.  Since  the  work  done  on  A  during  the  first  20  feet  =  758  foot-pounds, 
6.21  V2^  =  758,  or  V2  =  ii.o  feet  per  second. 

Such  a  problem  can  be  solved  also  by  first  finding  the  acceleration.  Thus, 
since  the  resultant  force  acting  on  ^  =  80  cos  20°  -  37.3  =  37.9  pounds,  the 
acceleration  =  37.9  ^  (400/32.2)  =  3.05  feet  per  second  per  second.  The 
time  for  describing  the  first  10  feet  =  the  velocity  acquired  -^  the  acceleration 
=  ^1/3-05  =  0-328  vi.    The  distance  =  the  average  velocity  X  the  time;  that 


Art.  43 


205 


Fig.  334 


is,  10  =  I  vi  X  0.328111,  or  Vi  =  7.81  feet  per  second  as  before.     Obviously, 
the  first  method  is  more  direct  than  the  second. 

2.  A  piece  of  timber  12"  X  12"  X  16'  is  suspended  by  means  of  two  parallel 
ropes  as  shown  in  position  A'B'  (Fig.  334).  The  ropes  are  10  feet  long  and 
the  timber  weighs  Soo  pounds.  It  is  raised  into  the 
position  AB,  two  feet  above  A'B',  and  then  allowed 
to  swing.  What  are  its  kinetic  energy  and  velocity 
when  it  reaches  its  lowest  position?  The  forces 
acting  on  the  timber  during  its  descent  are  gravity, 
the  pulls  of  the  ropes,  and  air  pressure.  We  neglect 
the  last.  At  each  instant  the  pulls  are  normal  to  the 
direction  of  the  displacement  of  their  respective  points  of  application;  there- 
fore the  pulls  do  no  work.  The  work  done  by  gravity  during  the  descent  = 
800  X  2  =  1600  foot-pounds.  Since  this  is  the  total  work  done  on  the  timber, 
the  kinetic  energy  of  the  timber  inUts  lowest  position  =  1600  foot-pounds. 
Now  the  timber  has  a  motion  of  translation  —  no  turning  — ,  and  therefore 
at  each  instant  all  points  of  the  timber  have  identical  velocities  (Art.  35). 
Hence,  ii  v  =  the  velocity  in  the  lowest  position,  then 

f  (800/32.2)  v~  =  1600,     or     D  =  11.35  feet  per  second. 

3.  A  certain  flywheel  and  its  shaft  weigh  400  pounds;  the  radius  of  gyration 
of  both  with  respect  to  the  axis  of  rotation  =  10  inches.  The  wheel  is  set  to 
rotating  at  100  revolutions  per  minute,  and  is  then  left  to  itself,  coming  to  rest 
under  the  influence  of  axle  friction  and  air  resistance  after  making  84  turns. 
Required,  the  average  torque  of  the  resistances.  The  moment  of  inertia  of 
the  wheel  and  shaft,  about  the  axis,  =  (400/32.2)  (10/12)^  =  8.64  slug-feet^. 
The  angular  velocity,  100  revolutions  per  minute,  =  2  tt  100/60  =  10.47 
radians  per  second.  Hence,  the  kinetic  energy  of  this  wheel  and  shaft,  when 
released,  =  |  8.64  X  10.47'  ~  474  foot-pounds.  Besides  the  forces  mentioned 
above,  gravity  and  the  normal  pressure  of  the  bearings  act  on  the  wheel  and 
shaft,  but  these  do  no  work  during  the  stoppage.  Let  M  =  average  torque 
of  the  resistances  in  foot-pounds;  then  the  work  done  by  them  during  the 
stoppage  is  —  M  2^84.  =  —528  M  foot-pounds.  This  equals  the  gain  in  the 
kinetic  energy  of  the  wheel;  that  is,  $28  M  =  474,  or  M  =  0.90  foot-pounds. 

4.  A  (Fig,  335)  is  a  sheave  supported  on  a  smooth  horizontal  shaft.  A  is 
3  feet  in  diameter,  and  its  radius  of  gyration  with  respect  to  the  axis  of  rotation 

=  9  inches.  The  weights  of  A,  B,  and  C  are  100,  200,  and  300 
pounds,  respectively.  The  system  is  released  and  allowed  to  move 
under  the  influence  of  gravity  and  the  resistances  brought  into 
action.  Required  the  velocity  of  the  suspended  weights  when 
they  have  moved  through  10  feet.  The  system  moves  under  the 
action  of  the  following  external  forces,  —  gravity,  axle  reaction, 
air  resistance,  and  the  internal  reactions  between  sheave  and  rope 
and  the  fibers  of  the   rope.      If  the  rope  is  quite   flexible  then  the  forces 


A 


C  5 

Fig.  335 


2o6 


Chap,  x 


in  the  rope  do  little  work;  this  will  be  neglected.  If  the  rope  does  not 
slip  on  the  sheave,  then  no  work  is  done  by  the  reaction  between  rope  and 
sheave.  Thus,  little  or  no  work  is  done  by  the  internal  forces.  The  work 
done  by  air  resistance  is  small  unless  the  speeds  of  the  moving  bodies  get  high; 
it  will  be  neglected.  The  work  done  by  the  frictional  component  of  the  axle 
reaction  per  turn  isil/2  tt,  where  M  is  the  frictional  moment  which  we  will  assume 
has  been  found  to  be  10  inch-pounds.  In  the  displacement  under  consideration, 
ID  feet  for  B  and  C,  the  wheel  makes  10/3  tt  turns.  Hence,  the  total  work 
done  by  friction  =  (10  X  2x)  (10/3  x)  =  66.7  inch-pounds  =  5.6  foot-pounds. 
Gravity  does  no  work  on  A;  or\.  B  and  C  its  work  =  300  X  10  -  200  X  10 
=  1000  foot-pounds.  We  neglect  its  work  on  the  rope  as  small.  Hence,  the 
total  work  done  on  the  system  =  1000  -  5.6  =  994.4  foot-pounds.  Now 
let  V  =  the  required  velocity  in  feet  per  second;  then  the  angular  velocity  of 
the  wheel  =.v-^  1.5  =  0.6677;  radians  per  second.  The  kinetic  energy  of 
the  system  equals 

I    300       „       .      I    200       „  I  .  N2 

-^5 — ■v~-\-- V -\- -I  {o.ob'jv)^, 

232.2  232.2  2 

where  7  =  moment  of  inertia  of  the  sheave.  Now  /  =  (100/32.2)  X  (9/12)^ 
=  1.75  slug-feet  2.  Hence,  the  kinetic  energy  of  the  system  =  8.16  z)^  foot- 
pounds.    Thus  the  work-energy  equation  is  994.4  =  8.16  v-;  hence  v  ^  11  feet 

per  second. 

5.  A  certain  pair  of  car  wheels  and  their  axle  weigh  2000  pounds.  Their 
diameter  is  33  inches  and  the  radius  of  gyration  of  wheels  and  axle  is  9  inches. 
They  are  rolled  along  a  level  track  until  their  speed  is  60  revolutions  per  min- 
ute, and  are  then  left  under  the  influence  of  the  rolling  resistance  of  the  track, 
coming  to  rest  after  rolling  a  distance  of  1000  feet.  (Data  not  from  an  actual 
experiment.)  Required,  the  average  rolling  resistance.  When  released,  the 
angular  velocity  of  the  wheels  =  one  revolution  per  second  =  6.28  radians 
per  second,  and  the  linear  velocity  of  their  centers  =  tt  33/12  =  8.64  feet  per 
second.     Hence,  the  kinetic  energy  = 

I   2000   _  „    ,     o      ,      I    2000 


X  8.64-  + (9/12)-  X  6.282  =  30J0  foot-pounds. 


2  32.2  2  32.2 

This  is  also  the  value  of  the  work  done  by  the  rolling  resistance,  air  resistance 
neglected.  Hence,  the  rolling  resistance  is  equivalent  to  a  constant  pull-back 
of  3010/1000  =  3  pounds. 

§  2.  Moving  Trains.  —  We  will  now  apply  the  principles  of  work  and 
energy  to  some  train  problems.  First,  we  briefly  consider  the  forces  directly 
concerned  with  the  motion  of  a  train  consisting  of  engine,  tender,  and  cars. 
For  convenience  we  regard  the  train  as  consisting  of  two  parts,  namely,  the 
locomotive  (engine  and  tender)  and  the  cars;   notation  as  in  Art.  42,  §  3. 

Locomotive.  —  For  simplicity  we  regard  the  locomotive  as  being  driven  by 
an  imaginary  (forward)  force  F  equivalent  to  the  steam  pressures.     To  be 


Art.  43 


207 


equivalent  the  work  done  by  F  per  unit  time  (or  power  of  F)  must  equal  the 
indicated  power  of  the  locomotive,  or  Fv  must  equal  pd^{l/D)  v;  hence, 

F  =  p  dH/D. 

This  force  F  we  will  call  the  cylinder  efort  of  the  locomotive.  The  resist- 
ances to  motion  experienced  by  a  locomotive  running  alone  on  a  straight 
and  level  track  may  be  put  into  three  groups:  —  (i)  Those  which  arise  through 
its  action  as  a  machine,  consisting  of  friction  in  the  working  mechanism 
(valves  and  gear,  cross-head,  piston,  crank  pins,  and  journals  of  driving-wheels) ; 
(ii)  those  which  arise  through  its  action  as  a  vehicle,  like  the  resistances  exper- 
ienced by  the  cars  (see  below) ;  (iii)  the  air  resistance.  For  convenience  we 
may  regard  all  the  resistances  in  each  group  lumped,  as  it  were,  into  a  single 
resistance  acting  backward  on  the  locomotive.  We  call  them  machine  resist- 
ance, vehicle  resistance,  and  frontal  resistance,  respectively;  and  we  desig- 
nate them  by  Rm,  Rv,  and  Rf.  The  sum  of  these  three  is  called  locomotive 
resistance,  and  will  be  denoted  by  Ri.  Thus,  we  regard  a  moving  locomotive 
as  under  the  action  of  the  following  forces  (see  Fig.  336) :  gravity,  the  support- 


T 


^ 


\        \ 


^J-^^r 


Fig.  336 


till 


ing  forces  of  the  track  (having  no  components  along  the  rails),  the  draw-bar 
pull  T,  the  locomotive  resistance  {Rm  +  -^w  +  Rf),  and  the  cylinder  effort  F. 
The  actual  external  forces  are  shown  in  Fig.  337. 


c 


y^ 


\ 


t   t      11 


n  n 


Fig.  337 


t       tilt 


If  the  velocity  of  the  locomotive  is  constant  and  the  track  is  straight  and 
level,  then  for  any  run  of  length  L  the  work-energy  equation  is 

[{p  dH/D)  -T  -  Ri]L  =  o;     hence     T  =  {p  dH/D)  -  Ri. 


2o8  Chap,  x 

If  the  velocity  is  changing,  then  the  power  equation  is 

where  v  =  velocity  of  locomotive  and  a  =  its  acceleration.     Hence 

T  =  ip  dH/D)  -  Ri-  Ma. 

If  the  locomotive  is  running  on  a  grade  then  the  grade  resistance  Rg  must  be 
included  in  an  obvious  way. 

According  to  the  American  Locomotive  Company  {Bulletin,  No.  looi),  the 
resistances  in  pounds  are  as  follows: 

Rf  =  0.24  V~,  where  V  is  velocity  in  miles  per  hour; 
R„  =  22.2  X  weight  on  drivers,  in  tons;  and 
R^,  =  the  same  as  for  cars  (see  further  on). 

The  Cars.  —  The  cars  are  urged  forward  by  only  one  force,  the  pull  of  the 
tender  on  the  first  car;  this  is  called  draw-bar  pull.  The  cars  are  retarded  by 
several  forces,  namely:  The  rolling  resistance  of  the  rails  upon  the  treads  of 
the  car  wheels;  the  journal  friction  at  the  axles  of  the  wheels;  the  air  resist- 
ance; and  miscellaneous  forces,  due  to  oscillation  and  concussion.  The 
"  laws"  of  these  separate  resistances  are  known  only  in  a  very  general  way. 
Because  of  lack  of  knowledge  of  these  separate  items  of  resistance,  and,  for 
convenience,  it  is  customary  to  "  lump"  them  into  a  single  equivalent  resist- 
ance, called  train  resistance.  Thus  we  may  imagine  trains  to  be  without 
actual  track,  journal,  air,  etc.,  resistance,  but  subjected  to  this  equivalent 
force,  conceived  as  a  single  pull  backward  on  the  train.  A  train  of  cars,  then, 
may  be  regarded  as  moving  under  the  action  of  four  forces,  namely,  the 
draw-bar  pull,  the  train  resistance,  gravity,  and  a  supporting  force  exerted  by 
the  track,  having  no  components  along  the  rails. 

Many  experiments  have  been  made  to  determine  train  resistance,  special 
"  dynamometer  cars"  (equipped  with  instruments  for  measuring  and  record- 
ing speed  of  train,  draw-bar  pull,  steam  pressure,  wind  velocity  and  direction, 
etc.)  being  used  for  that  purpose  now-a-days.  The  methods  for  determining 
train  resistance  are  very  simple  in  principle.  One  method  is  this:  —  the  loco- 
motive drags  the  cars  along  a  straight,  level  track  at  a  constant  speed;  the 
draw-bar  pull  and  the  speed  are  measured.  Then  the  (total)  train  resistance 
for  that  speed  equals  the  draw-bar  pull.  But  level  stretches  of  track  are  not 
always  convenient  of  access,  and  constant  speeds  are  not  easily  maintained. 
For  an  experiment  on  a  grade  let  H  =  the  ascent  or  descent  of  the  center  of 
gravity  of  the  train  during  the  experiment,  L  =  the  length  of  the  run,  W  = 
weight  of  cars,  T  =  average  draw-bar  pull,  R  =  average  train  resistance. 
Then  the  grade  resistance  is  Rg  =  ±WH/L,  according  as  the  train  i 
ascending  or  descending  the  grade,  and  the  work-energy  equation  is 

{T-Rt-  Rg)  L  =  E, 


IS 


Art.  43 


209 


where  E  is  the  gain  in  kinetic  energy  of  the  cars  during  the  run,  to  be  regarded 
as  negative  if  there  is  a  loss  of  kinetic  energy.     Hence 

Rt  =  T-Rg-  E/L. 

This  gives  average  train  resistance  for  the  speeds  of  the  run,  or,  perhaps,  the 
train  resistance  for  the  average  speed  of  the  run.  Another  method  is  based 
on  the  power  equation  (the  rate  at  which  work  is  done  on  the  cars  equals  the 
rate  at  which  they  gain  kinetic  energy);   this  is 


dt\2 


Mv^ )  =  Mva, 


where  M  =  mass  of  cars,  v  =  velocity,  and  a  =  acceleration.     Hence 

Rt  =  T  -  Rg-  Ma. 

If  the  train  is  being  retarded  then  a  should  be  regarded  as  negative.  There 
are  many  practical  difi&culties  in  carrying  out  experiments  as  suggested;  dis- 
cussion of  these  is  not  appropriate  here.* 

Obviously,  train  resistance  depends  upon  many  conditions,  as  state  of  track 
and  rolling-stock,  weather  and  wind,  and  velocity  of  train.  It  is  practically 
impossible  to  express  the  influence  of  all  these  conditions  in  a  formula  for  train 
resistance.     For  a  long  time  a  favorite  formula  was  the  so-called 

Engineering  News  formula,  r  =  2  +  j  F, 

where  r  =  train  resistance  in  pounds  per  ton  (weight  of  cars),  and  V  =  velocity 

of  train  in  miles  per  hour.     Recent  experiments  have  shown  very  clearly  that 

train  resistance  (per  ton)  depends  very  much  on  the  loading  of  the  cars,  being 

much   less   for   heavily   loaded   cars   than    for 

empties,  and  not  so  much  on  velocity  as  formerly 

belie\'ed.     The  American  Locomotive  Company 

in  Bulletin  No.  looi  states  that  "  The  best  data 

available  shows  that  the  resistance  varies  from 

about  2.5  to  3  pounds  for  72-ton  cars  to  6  to  8 

pounds  for  20-ton  cars"  (see  Fig.  338);  and  "for 

speeds  from  5  to  10  up  to  30  to  35  miles  per 

hour   the   resistance   is   practically   constant." 

Schmidt,  in  the  bulletin  already  mentioned,  gives 

formulas  for  train  resistance  (per  ton)  for  trains 

consisting   of   cars  of  different  average  weights;    also  the   following  as  an 

approximation 

_  V  -\-  39.6  —  0.031  w 
4.08  -{-  0.152  w 

where  r  =  train  resistance  in  pounds  per  ton,  V  =  velocity  in  miles  per  hour 
and  w  =  average  weight  of  cars  in  tons. 

*  See  Schmidt's  Freight  Train  Resistance,  University  of  Illinois  Bulletin  No.  43. 


20 


40        60 
Tons  per  Car. 

Fig.  338 


2IO  Chap.  X 

Examples.  —  i.  A  certain  locomotive  (engine  and  tender)  weighs  178.5  tons, 
106  tons  on  the  drivers.  There  are  two  cyhnders,  23  inches  (diameter)  X 
32  inches  (stroke);  the  drivers  are  63  inches  in  diameter;  and  boiler  pressure 
is  200  pounds  per  square  inch.  Required  the  maximum  draw-bar  pull  which 
this  locomotive  can  exert  on  a  level  track  at  20  miles  per  hour.  The  cylinder 
effort  is 

KpodH/D  =  {K  200  X  232  X  32)  -^  63  =  if  53,800  pounds. 

Now  the  piston  speed  s  =  2  vl/ir  D  (see  preceding  article)  =  (2  X  20  X  32) 
-^  (it  X  63)  =  6.465  miles  per  hour  =  569  feet  per  minute.  The  speed 
coefficient  (see  Fig.  331)  is  about  0.60;  hence  the  cylinder  effort  is  0.60  X 
53,800  =  32,300  pounds.  The  frontal  resistance  =  0.24  X  20'  =  96  pounds; 
the  machine  resistance  =  22.2  X  106  =  2350  pounds;  the  vehicle  resistance 
is  about  4  pounds  per  ton  or  4  X  (178.5  —  106)  =  290.  Hence,  the  total 
locomotive  resistance  is  about  2740  pounds,  and  the  maximum  draw-bar  pull 
=  32,300  —  2740  =  29,560  pounds  about. 

2.  A  freight  train  consists  of  30  cars,  average  weight  with  load  being  60  tons. 
What  is  the  "  resistance"  of  this  train  at  20  miles  per  hour?  According  to 
P.  R.  R.  (Fig.  338),  the  resistance  is  about  3.5  pounds  per  ton  or  6300  pounds 
total;  according  to  C.  B.  &  Q.,  it  is  about  2.5  pounds  per  ton  or  4500  pounds 
total.  According  to  Schmidt's  formula,  it  is  about  4.4  pounds  per  ton  or 
7920  pounds  total. 

3.  The  locomotive  (example  i)  pulls  the  train  (example  2)  along  a  straight 
track.  Required  to  show  graphically  how  the  cylinder  effort  and  the  various 
resistances  vary  with  the  velocity,  assuming  laws  of  resistances,  etc.,  as  in 
the  preceding  examples.  As  in  example  i  the  cylinder  effort  F=  iiT 53,800; 
the  piston  speed  5  =  2  vI/ttD  =  28.45  ^  ^^^t  per  minute  where  v  is  velocity  of 
locomotive  in  miles  per  hour.     Thus  we  have 


V  =        0 

5 

10 

15 

20 

25 

30 

35 

40 

s  =        0 

142 

284 

426 

569 

710 

853 

995 

1138 

K  =  o.&s 

•85 

.825 

•714 

.604 

•493 

•417 

•353 

.306 

F  =  22.9 

22.9 

22.2 

19.2 

16.2 

13.2 

II.  2 

9-5 

8.2 

the  value  of  K  having  been  taken  from  the  curve  A  BCD  (Fig.  331).  We 
next  compute  the  values  of  the  cylinder  effort  from  F  =  K  26.9  (tons)  for 
the  velocities  v.  Plotting  the  results  we  get  curve  i  (Fig.  339).  As  in 
example  i  we  take  machine  resistance  as  2350  pounds,  and  vehicle  resistance 
(for  locomotive)  as  290  pounds  at  all  velocities.  The  frontal  resistance  (given 
by  Rf  =  0.24  1)2)  varies.  At  d  =  o,  i?/  =  o;  and  at  z)  =  35  miles  per  hour, 
R^  =  385  about  pounds.  Inasmuch  as  R/  is  small  compared  with  the  other 
elements  of  locomotive  resistance,  we  will  take  a  mean  value  of  R/,  say  200 
pounds,  and  regard  it  as  correct  for  all  velocities.  Then  the  locomotive 
resistance  Ri  =  2350  -|-  290  -|-  200  =  2840  pounds.  Plotting  this  we  get 
the  straight  line  number  2.  Taking  train  resistance  at  3I  pounds  per  ton, 
as  in  example  2,  gives  Rt  =  6300  pounds,  assumed  to  be  independent  of 


fiRT.  44 


211 


velocity  for  the  range  from  5  to  35  miles  per  hour.     This  plotted  gives  line 
number  3. 

The  ordinate  between  lines  i  and  3  at  any  velocity  represents  the  net  or 
resultant  driving  or  accelerating  force  on  the  train  at  that  velocity.  Thus,  at 
20  miles  per  hour  that  ordinate  scales  about  23,500  pounds. 

Lbs. 
50  000 


40  000 


30  000 


20  000 


roooo 


^     / 

B 

^ 

•«^ 

^"^/-v 

^^ 

C 

D     ^ 

6300  lbs 

■3 

f 

iRi=  284-0  lbs\ 

10 


20 


E5 


30 


35 


Miles  per  Hour 
Fig.  339 

4.  Referring  to  the  train  of  the  preceding  example:  —  Required  to  show 
how  its  acceleration  varies  with  the  velocity.  Under  the  preceding  example 
it  was  explained  that  any  ordinate  between  lines  i  and  3  represents  net  driving 
force  on  the  train.  Hence  the  acceleration  at  any  velocity  =  such  ordinate 
(to  scale)  divided  by  the  mass  of  the  train.  Thus  at  20  miles  per  hour,  say,  the 
ordinate  represents  about  23,500  pounds,  and  the  acceleration  =  23,500  4- 
(3^957^00/32.2)  =  0.1914  feet  per  second  per  second.  In  this  way  the 
accelerations  at  other  velocities  might  be  computed,  and  then  the  curve 
of  accelerations  determined.  This  curve  would  resemble  curve  i;  indeed 
the  accelerations  are  proportional  to  the  ordinates  from  line  3  to  line  i. 


44.   Efficiency;  Hoists 

§  I.  Efficiency  of  Machines.  —  Among  the  machines  and  appliances 
used  in  the  industries  there  are  some  whose  function  is  the  conversion  or 
transmission  of  energy.  For  example,  —  an  electric  dynamo  which  converts 
mechanical  into  electrical  energy;  a  steam  engine  which  converts  energy  of 
steam  into  (kinetic)  energy  of  its  flywheel;  a  line-shaft  which  transmits 
energy  from  one  place  in  a  shop  to  one  or  more  other  places;  etc.  In  this 
article,  "machine  "  means  the  kind  of  machine  or  appliance  just  described.  The 
amount  of  energy  supplied  to  a  machine  in  any  interval  of  time,  for  conversion 
or  transmission,  is  called  the  input  for  that  time;  the  amount  of  energy  con- 
verted into  the  desired  form  or  transmitted  to  the  desired  place  is  called  the 
output.     Experience  has  shown  that  output  is  always  less  than  input;    that 


212  Chap.x 

is  a  machine  does  not  convert  or  transmit  the  entire  input.  The  difiference 
between  output  and  input,  for  the  same  interval  of  time  of  course,  is  called 
lost  energy  or  loss  simply.  By  efficiency,  in  this  connection,  is  meant  the  ratio 
of  output  to  input;   that  is  if  e  =  efficiency,  then 

e  —  output  -i-  input. 

Most  machines  are  designed  for  a  definite  rate  of  working  or  for  a  certain  load 
called  its  full  load.  Then  we  speak  of  the  efficiency  of  a  machine  at  full  load, 
half-load,  quarter  over-load,  etc.,  these  efficiencies  being  different  generally. 
The  two  following  tables  are  given  to  furnish  some  notion  of  the  efiiciencies 
of  the  more  common  machines.* 


Full-load  Efficiency  of  Efficiency  of  Some  Machine  Elements* 


— Per  cent  P'^'"  cent 

Hydraulic  turbines 60-85      Common  bearing,  singly 96-98 

impulse  wheels     75-85      Common  bearmg,  long  Imes  of  shaftmg. . .  95 

Roller  bearings 9^ 

Steam  boilers 50-75      Ballbearings ..         .. 99 

engines 5-20      Spur  gear  cast  teeth,  mcludmg  beanngs.  .  93 

turbines .  . 5-20      Spear  gear  cut  teeth,  including  bearings.  .  96 

Bevel  gear  cast  teeth,  including  bearings.  92 

Gas  and  oil  engines 16-30      Bevel  gear  cut  teeth,  including  bearings. .  95 

Belting 96-98 

Electric  dynamos 80-92       Pin-connected  chains,  as  used  on  bicycles.  95-97 

motors 75-90      High-grade  transmission  chains 97-99 

transformers. .  . .     50-95 


*  From  Kimball  and  Barr's  Elements  of  Machine  Design. 

The  efficiency  of  a  combination  of  machines.  A,  B,  C,  etc.,  A  transmitting 
to  B,  B  toC,  etc.,  is  the  product  of  the  efficiencies  of  the  individual  machines. 
For,  let  61,  €2,  ez,  etc.  =  the  efficiencies  of  the  separate  machines  A,B,C,  etc., 
and  e  =  the  efficiency  of  the  group.  Then  if  £  =  the  input  for  A,  the  output 
of  ^  =  exE  =  the  input  for  B;  the  output  oi  B  =  CieiE  =  the  input  for  C; 
the  output  of  C  =  eze^eiE  =  the  input  for  D;  etc.  Hence,  the  output  of  the 
last  machine  -^  the  input  of  the  first  =  {eie^ez  .  .  .  )E  -^  E  =  exe^ez  .  .  .  ,  or 

e  =  e\' ei' ez  .  .  .  . 

For  example,  if  a  dynamo  is  run  by  a  steam  engine,  then  the  efficiency  of  the 
combination  or  set  =  the  product  of  their  separate  efficiencies,  say  0.20  X 
0.90  =  0.18  or  18  per  cent. 

§  2.  Hoisting  Appliances,  Etc.  —  There  are  certain  rather  simple  ap- 
pHances  by  means  of  which  a  given  force  can  overcome  a  relatively  large 
resistance;  as,  for  example,  the  lever,  the  wedge,  the  screw,  the  pulley,  etc. 
Such  an  appliance  is  generally  operated  by  means  of  a  single  force,  which  we 

*  For  detailed  information  see  IMead's  Water  Power  Engineering,  from  which  most  values 
in  the  first  table  were  taken;  Gebhardt's  Steam  Power  Plant  Engineering;  and  Franklin  and 
Esty's  Elements  of  Electrical  Engineering. 


Art.  44  ^•'■3 

call  driving  force  and  denote  by  F;  it  is  called  effort  also.  The  force  which 
the  appliance  is  desired  to  overcome  we  call  resisting  force  and  denote  by  R, 
or  when  the  force  is  a  weight,  by  t^;  it  is  called  resistance  also.  In  many 
appliances  (hereafter  called  "common  ")  all  equal  displacements  of  the  point 
of  application  of  the  driving  force  result  in  equal  displacements  of  the  point 
of  application  of  the  resisting  force;  and  generally  these  displacements  re- 
spectively take  place  along  the  lines  of  action  of  the  driving  and  resisting 
forces.  These  displacements,  or  their  components  along  the  lines  of  action 
of  the  forces  respectively  if  the  displacements  are  inclined  to  the  forces,  we 
will  denote  by  a  and  b  respectively. 

In  a  common  appliance,  the  work  done  by  the  driving  force  and  (by  the 
appUance)  against  the  resisting  force  are  respectively  Fa  and  Rb;  hence  the 

efficiency  is  given  by 

e^  Rb^  Fa.  (i) 

Let  Fo  =  the  effort  which  would  be  required  to  overcome  the  resistance  R 

if  the  machine  were  frictionless;  then  Foa  =  Rb.     Substituting  in  (i)  we  find 

that  efficiency  is  given  also  by 

e  =  Fo-^F.  (i') 

Let  Ro  =  the  resistance  which  F  could  overcome  if  the  machine  were  friction- 
less;    then  Fa  =  Rob.     Substituting  in  (i)  we  find  that  efficiency  is  given 

also  by 

e  =  R-^Re.  (i  ) 

Most  of  the  appliances  now  under  discussion  can  be  operated  backward  as 
well  as  directly.  For  example,  the  lever,  the  wedge,  the  screw,  etc.,  can  be 
used  to  lower  a  heavy  body  as  well  as  to  raise  it.  Some  of  these  appliances, 
which  can  be  run  either  way,  will  run  backward  without  direct  assistance 
when  loaded;  that  is  the  load  will  overcome  the  internal  friction.  Such 
appliances  are  said  to  overhaul.  Some  will  not  run  backward  unassisted; 
that  is  the  load  cannot  overcome  the  internal  friction.  Such  appUances  are 
said  to  self-lock.  An  appliance  will  overhaul  or  self-lock  according  as  its 
(direct)  *  efficiency  is  greater  or  less  than  one-half,  if  the  works  done  in 
overcoming  friction  in  a  forward  and  in  an  equal  backward  motion  are  equal 
(usual  case).  Proof:  — As  before,  let  F  =  the  effort,  R  =  the  (useful)  re- 
sistance, a  and  b  =  corresponding  displacements  of  F  and  R,  and  w  =  the 
work  done  against  friction,  all  in  forward  motion  of  the  appliance.  Then 
Fa  =  Rb  -\-  w.  Now  if  the  efficiency  (forward  motion)  is  greater  than  one- 
half,  then  more  than  one-half  of  the  work  Fa  is  expended  usefully  (against 
R) ;  that  is  Rb  is  greater  than  w,  and  hence  R  could  overcome  the  friction  in 
backward  motion.     If  the  efficiency  (forward  motion)  is  less  than  one-half. 

*  When  a  machine  is  run  backwards  it  is  said  to  have  reversed  efficiency,  by  (considerable) 
extension  of  the  definition  of  efficiency.  In  such  case  the  load  (on  the  hoist,  for  example) 
is  the  effort,  and  the  applied  force  is  regarded  as  the  useful  resistance.  In  case  the  machine 
selflocks  so  that  the  applied  force  (P  say)  must  assist  the  load,  then  by  considerable  stretch 
of  imagination  —  P  is  regarded  as  the  useful  resistance;  the  computed  (reversed)  efficiency 
is  negative. 


214  Chap,  x 

then  less  one-half  the  work  Fa  is  done  against  R;  that  is  Rh  is  less  than  w, 
and  hence  R  could  not  overcome  friction  unassisted  in  backward  motion. 

By  mechanical  advantage  of  an  appliance  is  meant  the  ratio  of  the  resisting 
to  the  driving  force  when  the  appliance  is  operating  steadily,  at  constant 
speed.    Thus,  see  equation  (i),  mechanical  advantage  is  given  by 

R/F  =  e  a/b  (2) 

Obviously  the  value  of  the  ratio  a/h  does  not  depend  on  the  loss  or  wasted 
work;  that  is,  it  is  independent  of  the  efficiency.  (The  ratio  depends  solely 
on  the  geometrical  proportions  of  the  appliance.)*  Hence  we  may  assume 
e  =  I  and  write  a/b  =  R'/F'  where  R'/F'  means  the  mechanical  advantage 
of  the  appUance  if  it  were  without  friction.     Finally, 

R/F  =  e  R'/F'  (2') 

or,  (mechanical  advantage)  =  (efficiency  c)  X  (mechanical  advantage  at  e  =1). 
In  some  appliances  or  mechanisms  the  driving  and  (useful)  resisting  forces 
{F  and  R)  are  applied  at  a  wheel  (pulley,  gear,  etc.),  and  it  is  more  convenient 
in  the  discussion  to  deal  with  the  torques,  or  moments  of  the  forces  about  the 
shaft  axes  respectively,  than  with  the  forces.  Let  Ti  and  T2  denote  those 
torques,  of  the  driving  and  resisting  forces  respectively,  and  a  and  /5  corre- 
sponding angular  displacements,  in  radians,  of  the  wheels  to  which  Ti  and  T2 
are  applied.  The  works  done  by  the  force  F  and  against  R  during  the  dis- 
placements a  and  /3  are  Tia  and  T^^.  Hence,  the  efficiency  is  e  =  T-S/Txa, 
and 

Ts/ri  =  ea/^.  (3) 

Reasoning  as  in  the  preceding  paragraph  we  conclude  that  a/jS  =  T2/T1  where 
T2  /Ti  means  the  ratio  of  the  resisting  torque  to  the  driv- 
ing torque  if  the  appliance  were  frictionless,  e  =  i.     Hence 

T,/T,  =  e  T^'/T,'.  (3') 

We  may  call  the  ratio  of  the  resisting  and  driving  torques, 
the  mechanical  advantage  of  torque;  then  the  foregoing 
result  may  be  stated  as  follows: 

(mechanical  advantage  of  torque)  =  (efficiency  e)  X  (me- 
chanical advantage  of  torque  at  e  =  i). 

Examples.  —  i.  The  pitch  of  the  screw-jack  (Fig.  340)  is 
h,  the  mean  radius  of  the  screw  thread  is  r,  the  length  of 
the  lever  is  /.  What  is  the  efficiency  of  the  jack  when  it  is 
overcoming  (raising)  IF?  It  is  shown  in  Art.  20  that  the  force  required  at 
the  end  of  the  lever  to  start  the  screw  is  P  ==^  W  (r/l)  tan   (^  +  a),  where 

*  When  the  displacements  a  and  b  are  not  inclined  to  the  forces  F  and  R  respectively, 
then  the  ratio  a/b  is  sometimes  called  the  velocity  ratio  of  the  appliance,  for  the  velocities 
of  the  points  of  apphcation  of  F  and  R  are  as  a  to  b.     Thus  we  have  for  such  cases 

mechanical  advantage  =  efficiency  X  velocity  ratio. 
High  mechanical  advantage  requires  high  velocity  ratio,  b  small  compared  to  a;   thus  the 
adage  "what  is  gained  in  force  is  lost  in  velocity." 


Art.  44 


215 


Fig.  341 


(f)  is  the  angle  of  static  friction  and  a  is  the  pitch  angle,  tan~^  {h/2  irr). 
Hence  the  force  required  to  raise  the  load  W  uniformly  is  given  by  the  same 
expression  if  0  is  taken  to  denote  the  angle  of  kinetic  friction  (see  Art.  45). 
Let  it  be  so  understood  in  what  follows.  If  the  screw  were  frictionless  then 
the  force  required  would  be  Po  =  W  (r/l)  tan  a;  and  hence  the  efficiency  is 
e  =  tan  a  -^  tan  ((^  +  a). 

2.  Fig.  341  represents  a  double  purchase  crab,  for  hoisting.  Hand  cranks 
can  be  applied  on  the  ends  of  the  shaft  B  or  C  The  hoisting  rope  winds  on 
the  drum.  The  crank  is  18  inches  long;  the  drum  is  10  and  the  rope  f  inches 
in  diameter;  gears  A'  and  B  are  20  and  B'  and  C 
are  4  inches  in  diameter.  Required  the  mechanical 
advantage  of  the  appliance  when  a  crank  is  used 
on  C.  Evidently,  angular  displacements  of  gears  A ' 
and  B'  respectively  are  as  i  to  5 ;  also  the  angular 
displacements  of  gears  B  and  C.  Hence,  for  one 
turn  of  A',  C  makes  5  X  5  ==  25  turns.  For  one 
turn  of  A',  b  =  10.75  tt,  ^.nd  a  (the  corresponding 
displacement  of  the  point  of  application  of  the 
effort)  =  36X  X  25  =  900 TT.  Therefore,  the  velocity  ratio  of  the  appliance 
is  83.8.  If  the  efficiency  is  So  per  cent  say,  the  mechanical  advantage  is 
0.80  X  83.8  =  67.     See  equation  (2). 

This  result  could  be  obtained  from  equation  (2')  as  follows:  Let  P'  = 
effort  at  the  crank  handle,  T'  =  tooth  pressure  between  gears  ^-I'  and  B', 
T  =  tooth   pressure  between   gears  B   and  C,  6'  =  obliquity  *    of    T',   d  = 

*  Fig.  342  represents  parts  of  two  gears  Oi  and  O2. 
Imagine  the  tops  of  the  teeth  cut  ofi  and  the  remaining 
hollows  filled  in  so  that  the  toothed  wheels  become  friction 
wheels,  transmitting  power  by  friction  (developed  by  press- 
ing the  wheels  together).  Evidently  the  diameters  of  these 
(imaginar\')  friction  wheels  could  be  chosen  so  that  the  ratio 
i  of  the  angular  velocities  of  the  wheels  would  be  equal  to 

/  that  of  the  gears.     The  intersections  of  the  faces  of  such 

wheels,  of  the  particular  diameters  mentioned,  and  a  plane 
perpendicular  to  the  shafts  are  the  "pitch  circles"  of  the 
gears.  The  common  point  of  the  two  pitch  circles  is  the 
Q  "pitch  point."  To  insure  constant  ratio  of  angular  ve- 
locities of  two  gears  in  mesh,  the  form  of  the  teeth  must 
be  such  that  the  normal  to  the  surfaces  of  two  teeth  where 
they  touch  passes  through  the  pitch  point.  (For  proof, 
see  any  standard  book  on  Mechanism.)  The  angle  be- 
tween this  normal  and  the  common  tangent  of  the  pitch 
circles  is  called  "  obliquity  "  of  the  normal.  In  some  gears 
this  obliquity  remains  constant,  as  the  gears  turn;  in 
others,  it  varies.  In  the  figure,  0  is  the  pitch  point,  OT 
the  common  tangent  to  the  pitch  circles,  AO  the  normal 
for  the  teeth  touching  at  A ,  and  BO  the  normal  for  the 
teeth  touching  at  B;  the  obliquity  of  these  normals  are  denoted  by  0'  and  d". 
If  the  teeth  were  frictionless,  the  pressure  between  two  teeth  (in  contact)  would  act  along 


Fig.  342 


2l6 


Chap,  x 


obliquity  of  T,  and  R'  =  resistance,  —  all  on  the  supposition  that  the 
appliance  is  frictionless,  or  e  =  i.  Then,  considering  torques  on  the  three 
shafts  it  is  plain  that 

P'  X  i8  =  r  X  2  cos  ^,    r  X  lo  cos  ^  =  r'  X  2  cos  6', 
r  X  locos^'  =  R'  X  5.375; 
and  from   these  it  follows  that  R'/P',  the  mechanical  advantage  at  e  =  i, 
is  83.8.     Hence  the  (true)  mechanical  advantage  is,  as  before,  83.8  X  0.80  =  67. 
3.    Fig.  343  shows  the  operating  mechanism  for  a  lift-bridge,  but  there  is 
/■Tower  posf  of  fixed  span 


Truss   of  liff  span 


ffack 


Equalizing  sinaff  '^0  gearing  at 

opposite  end  of  span 

Socket  for.,     iL-A 
capstan      ^  ^  ^TZc^f^FTi^  IThfi  s^ 

T-.'  •  IT    r^ 

z. 


Truss  of  lift  span 


'\       |<- Liff  span 

''■■Tower  post  of  fixed  span 

Fig.  343 

the  normal,  at  the  contact  pohit,  and  hence  through  the  pitch  point.  Let  TV  denote  the 
resultant  of  all  the  pressures  on  the  frictionless  teeth  of  either  gear;  obviously  iV  passes 
through  0.  Imagine  iV  resolved  at  O  into  components  along  and  perf)endicular  to  OT,  and 
let  N  t  denote  the  first  component.  Then  the  (frictionless)  torques  exerted  by  the  driving 
and  driven  gears  on  each  other  are  respectively 

iVt  I  dx  and  A"t  \  d-i,  (i) 

where  d\  and  dt  are  the  diameters  of  the  pitch  circles  of  the  gears  respectively. 

On  account  of  the  tooth  friction,  each  tooth  pressure  is  inclined  to  the  normal  at  the 
contact,  by  an  amount  equal  to  the  angle  of  friction  <^  (see  Art.  45).  At  two  teeth  which  are 
approaching,  the  pair  touching  at  A,  the  directions  of  the  frictional  forces  are  such  that., 
the  obliquity  (to  OT)  of  the  tooth  pressure  R'  is  Q'  +  4>.  At  two  teeth  which  are  receding, 
the  pair  touching  at  B,  the  directions  of  the  frictions  are  such  that  the  obliquity  (to  OT)  of 
the  tooth  pressure  R"  is  B"  —  <^.  In  either  case,  the  Hne  of  action  of  the  tooth  pressure 
cuts  the  line  joining  the  centers  of  the  gears  between  the  pitch  point  O  and  the  center  O2 
of  the  driven  gear.  It  follows  that  the  line  of  action  of  the  resultant  of  all  the  tooth  pres- 
sures on  either  gear  cuts  the  line  O1O2  in  a  point  C  between  0  and  O2. 

Let  .V  denote  the  distance  from  C  to  0,  R  the  mentioned  resultant,  and  Rt  the  compo- 
nent of  R  parallel  to  OT.  Then  the  torque  exerted  by  the  driving  and  the  driven  gears 
on  each  other  are  respectively 

Rtildi-x)      and      Rtihdi+x). 
If  e'  is  the  efficiency  of  the  gears  alone,  shafts  frictionless,  then 

did-2 


(i  —  e'))  approximately. 


(2) 


2  di  -\-  d2 

For,  let  Ti  be  a  torque  applied  to  drive  the  driving  gear,  and  Tj  a  resisting  torque  applied  to 
the  driven  gear;  then 

e'  =  Tsrfi  ^  T,d2,       Ti  =  Rt  ih  di  +  x),       T-i  =  Rt  {h  d.  -  x). 
These  three  equations  yield  (2). 


Art.  44  217 

a  duplicate  mechanism  on  the  other  end  of  the  Hft  span  not  shown.  All 
this  mechanism  rests  on  and  moves  with  the  lift  span.  There  are  fixed  tower 
posts  adjacent  to  the  lift  span  as  shown,  and  on  these  posts  there  are  fixed 
vertical  racks  which  engage  spur  pinions  GG  of  the  operating  mechanism. 
The  pinions  are  driven  by  the  motors  or  hand  capstans,  and  thus  the  Hft  span 
is  raised  or  lowered.     The  intermediate  drive  consists  of: 

a  motor  pinion  A number  of  teeth,  18,      diameter,     5 .  14  inches 

cross  shaft  gear  B 126  36.00 

cross  shaft  bevel  pinions  CC 20  10 .00 

counter  shaft  bevel  gears  DD 60  30.00 

counter  shaft  spur  pinions  EE 15  8.36 

operating  shaft  spur  gear  FF 52  29 .00 

operating  shaft  pinion  GG 15  11  -94 

bevel  pinion  on  hand  operator  A' 16  12.72 

bevel  gear  on  hand  operator  B' 24  19.08 

The  lift  span  is  accurately  counterweighted  so  that  no  work  is  done  against 
gravity  either  in  raising  or  lowering;  but  vertical  pressure  must  be  developed 
between  the  racks  and  the  pinions  G  to  overcome  the  (internal)  friction  in  the 
counter  weight  mechanism.  It  is  estimated  that  a  vertical  pressure  of  5000 
pounds  is  required  at  each  pinion  for  lifting  the  bridge.  How  great  a  driving 
torque,  at  each  motor,  is  required  to  develop  this  pressure  at  each  pinion  G? 

We  will  first  neglect  losses  in  mechanism.     Let  2  T/  be  the  driving  torque 
at  the  motor  and  T2'  the  resisting  torque  at  the  pinion  G.     Then  the 

torque  to  the  cross  shaft  =  2  T/  -j%^-, 
torque  to  the  counter  shaft  =  Ti  ~\%^  |§, 
and  torque  to  the  operating  shaft  =  Ti'  VV'  In  tI  =  36.9  T/. 

But  36.9  r/  =  T2',  or  T2  jTi  =  36.9.  We  take  the  efficiencies  of  the  gears 
to  be  as  follows:  A  and  5,  96  per  cent;  C  and  D,  92  per  cent;  E  and  F,  96 
per  cent.  Then  the  efficiency  of  the  transmission  from  the  motor  to  the 
pinion  G  is  0.96  X  0.92  X  0.96  =  0.85,  or  85  per  cent.  Hence  the  ratio  of 
the  actual  driving  and  resisting  torques  is  T^lT^  =  0.85  X  36.9  =  31.4.  The 
lever  arm  of  the  vertical  pressure  of  the  rack  against  the  pinion  G  is  5.97  +  a; 
(see  footnote  on  page  216).  Since  di  (diameter  of  the  pitch  circle  of  the  rack) 
is  infinite,  equation  (2)  as  written  is  not  usable;   it  can  be  written 

which,  since  d\l di  =  o  in  the  present  problem,  becomes 


% 


_  1 

~  2 


di  (i  —  e')  =  2  X  11.94  (i  —  0.98,  say)  =  0.12  inches. 


Therefore,  the  actual  resisting  torque  on  the  pinion  is 

5000  X  6.09  =  30,450  inch-pounds  =  2537  foot-pounds. 

Hence  Ti  =  2537  -^  31.4  =  80  foot-pounds,  and  the  required  motor  torque 
is  160  foot-pounds. 


2l8 


Chap,  x 


-^ 

/ 

*  i 

:^ 

^ 

\ 

1    J 

^ 

V 

^ 

^ 

1      M 

U...R-.J 

u 

'       T| 

Fig.  344 


Pulley.  —  Fig.  344  represents  a  simple  pulley  with  part  of  a  rope  or  chain 
upon  it.  Let  S  =  tension  in  leading  or  off  side  of  the  rope,  and  T  =  tension 
in  the  following  or  on  side.  Then  evidently  S  is  greater 
than  r,  for  5  overcomes  not  only  T  but  also  the  friction 
at  the  pin  and  the  "rigidity"  of  the  rope.  The  resistance 
due  to  pin  friction  =  the  product  of  the  coefiScient  of 
axle  friction  (see  Art.  45)  and  the  pressure  on  the  pin; 
this  pressure  =  S  -\-  T.  Hence,  if  /  =  coefficient  and 
r  =  radius  of  the  pin,  the  work  done  against  friction  per 
revolution  of  the  pulley  =  2  tt  r/  (5  +  T).  The  work  done 
in  bending  or  unbending  (inelastic)  rope  over  the  pulley 
is  proportional  to  the  amount  of  rope  so  bent  per  revolution  (that  is  2  ttR), 
and  it  seems  to  be  proportional  also  to  the  tension,  to  the  area  of  the  cross 
section  of  the  rope  and  inversely  proportional  to  the  radius  R.  Thus  the 
work  of  bending  =  C2TrRT  d^/R  =  CnrTd'-,  where  d  =  diameter  of  the  rope 
and  C  is  an  experimental  coefficient  depending  on  the  kind  of  rope  and  perhaps 
other  elements.  Likewise  the  work  of  unbending  (at  off  side)  =  CiirSd^. 
Now,  if  we  equate  the  work  done  by  the  effort  5  to  the  work  done  against 
rigidity,  the  resistance  T,  and  the  axle  friction,  and  then  simplify  the  resulting 
equation,  we  get        /  .  j2n       ^/     ,    ,r    ,   ^J2\ 

^V--^r-^r)^^V^-^r^^r} 

This  equation  can  be  written  in  the  following  approximately  correct  form,  — 

5  =  (i  +  2/^  +  2c|)r  =  irr, 

where  K  is  an  abbreviation  for  i  +  2  fr/R  +  2  C  d^/R.  According  to  experi- 
ments by  Eytelwein,  C  equals  about  0.23  when  d  and  R  are  expressed  in  inches. 
The  American  Bridge  Company  made  some  experiments  to  determine  C  and 
K  for  such  pulleys  and  rope  as  are  in  common  use  in  tackle  for  construction 
work,  and  found  that  C  depends  not  only  on  kind  of  rope,  as  expected,  but  also 
on  the  size  of  rope.     The  following  table  is  taken  from  their  report.* 


Dimensions  and  coefficients 

Hemp 

Wire 

Diameter  of  rooe,  d   

li             I§             if             2 

3I        4A       5l        6 

i         I           li         if 
0.23     0.20    0.19     0.17 
1.20     I. 21     1.23     1.24 

3 

Center  pin  to  center  rope,  R 

7l 

Diameter  of  oin.  2  /* 

2h 

Coefficient  C                         

0.9 

Coefficient  K 

1. 16 

These  values  of  C  and  K  are  higher  than  those  usually  employed.  C  =  0.08 
to  0.22  is  advised  for  hemp  rope,t  and  "  K  =  1.06  to  1.07  may  be  considered 
maximum  practical  values  since  generally  K  =  1.02  to  1.04."^ 

*  Trans.  Am.  Soc.  C.  E.,  1903,  Vol.  51,  p.  161;   also  Eng.  Rec,  1903,  Vol.  48,  p.  307. 

t  Hiitte,  Taschenhuch  (Twentieth  Edition),  Vol.  i,  p.  247. 

X  Bottcher-Tolhausen,  Cranes,  p.  15. 


Art.  44 


219 


Tackle.— When  a  fixed  pulley  (Fig.  345),  for  example,  is  used  for  lift- 
ing, P  =  KW  or  W/P  =  i/K;  for  lowering,  W  =  KP,  or  P/IV  =  i/K.  When 
a  movable  pulley  (Fig.  346)  is  used  for  lifting,  IF  =  P  +  T  =  P  +  P/K,  or 
W/P  =  (i  +  K)/K;  for  lowering,  W  =  P -\-  S  =  P -\- KP,  or  W/P  =  i  +  A'. 

In  a  similar  way  we  can  determine  the  mechanical  advantage  of  any  com- 
bination of  pulleys  in  terms  of  K.  For  example,  consider  the  tackle  repre- 
sented in  Fig.  347.     There  are  two  separate  pulleys  in  each  block  A   and 


''/(//////////{^i     -'//(///////////^i, 


Fig.  345 


Fig.  346 


B.  The  pulleys  in  a  block  are  generally  alike  in  size  but  are  here  represented 
unlike  for  clearness.  Let  P  =  the  applied  pull  and  W  =  load.  Pi,  P2,  P?. 
and  P4  =  the  tensions  as  indicated  in  Fig.  348.  When  the  tackle  is  used  for 
lifting.  Pi  =  P/K,  P2  =  Pi/K  =  P/K\  P3  =  PoJK  =  P/X^  and  P4  = 
P3/K  =  P/K\  And  since  PF  =  Pi  -^  P2  +  P3  +  P4,  we  have  also 
P       P       P        P       P 


or 


W/P  =  (A'3  -f  K^  +  A'  -f  i)/A4. 
When  the  tackle  (Fig.  347)  is  used  to  lower  the  load, 

Pi  =  A"P,  P2  =  /CPi  =  K^P,  Pz  =  KP^  -  A^P,  Pi  =  KPi  =  A'4P. 


-/////////////////////.    -//////////////////■ 


w//'/jum>     ■w////(////M. 


Fig.  347 
And  since 


Uw     !  jw 


W 

Fig.  348 

TF  =  Pi  -f  P2  +  P3  +  P4, 
we  have     TF  =  AP  +  A^P  +  A''P  +  A^P  =  P  (A  -j-  A^  -f  K^  -\-  K^) 
or  W/P  =  A  (i  -F  A  +  A2  +  A^). 


220  Chap,  x 

Special  {Chain)  Hoists.  —  Fig.  349  represents  a  Weston  differential  hoist. 
The  upper  block  contains  two  pulleys  differing  sUghtly  in  diameter;  they  are 
fastened  together.  The  lower  block  contains  only  one  pulley.  The  pulley 
grooves  have  pockets  into  which  the  links  of  the  chain  fit;  thus  slipping 
of  the  chain  is  prevented.  The  chain  is  endless  and  is  reeved  as  shown.  If 
there  were  no  lost  work,  then  the  tension  in  each  portion  of  the  chain  to  block 
B  would  equal  one-half  the  load  (Fig.  350),  and  the  pulls  on  the  block  A  would 
be  as  indicated  in  the  figure.  Now  if  R  and  r  =  the  distances  from  the  center 
of  the  pin  in  block  A  to  the  axis  of  the  chain  as  indicated  then  moments  about 
the  axis  of  the  pin  give 

PoR  +  h  ^Vr  =  \  WR,     or     W  =  Pq  2  RliR  -  r)\ 

the  ratio,  HV^o  =  2R/{R  —  r)  may  be  made  very  large  by  making  R  —  r  small. 
The  mechanical  advantage  is 

IF       W  2  R 

e  = 


P       Po        R-r' 

where  P  =  the  actual  force  required  to  raise  W  and  e  =  efficiency.  These 
hoists  are  made  of  various  capacities  up  to  W  =  3  tons;  their  efficiencies  are 
relatively  low,  from  about  25  to  40  per  cent  according  to  the  manufactiu-ers' 
lists.  In  the  so-called  Duplex  and  Triplex  hoists  the  upper  blocks  are  screw- 
geared  and  spur-geared  respectively.  At  full  load  the  efficiency  of  these 
hoists  varies  from  about  30  to  40  and  from  70  to  80  per  cent. 

Example.  —  We  will  now  show  how  to  apply  some  of  the  preceding  prin- 
ciples and  formulas  in  a  computation  relating  to  the  operating  machinery  of 
the  vertical  lift  bridge  represented  in  Figs.  351  and  352.  The  Hft  span  when 
down  in  place  rests  on  two  piers.  When  up  it  is  balanced  by  two  counter- 
weights as  shown.  Each  counterweight  is  suspended  by  means  of  two  pairs 
of  one-inch  cable;  each  pair  of  cables  extends  upwards  from  the  counterweight, 
over  a  sheave  and  downward  to  a  point  of  attachment  on  the  lift  span.  At 
each  corner  of  the  lift  span  there  is  a  spirally  grooved  drum  carrying  two  one- 
half-inch  cables.  Each  cable  has  one  end  attached  to  its  drum;  the  other  end 
of  the  up-haul  cable  is  attached  to  a  point  vertically  above  at  the  top  of  the 
tower,  and  the  other  end  of  the  down-haul  cable  is  similarly  attached  at  the 
base  of  the  tower.  As  the  drum  is  revolved,  one  cable  is  wound  upon  it  and 
the  other  is  paid  out.  The  tw^o  drums  at  either  end  of  the  span  are  mounted 
upon  a  single  cross-shaft  A,  which  carries  a  bevel  gear  B.  The  gears  BB  mesh 
with  bevel  pinions  DD  mounted  on  the  longitudinal  shaft  C  which  also  carries 
a  bevel  gear  E.  E  meshes  with  a  bevel  pinion  F  on  a  vertical  shaft  which 
carries  a  capstan  head.  This  capstan  head  takes  a  horizontal  lever  by  means 
of  which  a  man  operates  the  mechanism.  To  lift  the  bridge  he  rotates  the 
capstan  headed  shaft  in  the  proper  direction  and  drives  the  drums;  they  wind 
the  up-haul  cable  upon  themselves  and  pay  out  the  down-haul  cable  as  already' 
described.     This  winding  up  necessitates  upward  motion  of  the  bridge. 


Art.  44 


221 


The  length  of  the  lever  (radius  of  circle  in  which  the  man  walks  as  he  oper- 
ates) is  6  feet.  The  pinions  F  and  D  are  alike;  each  is  6.86  inches  in  diameter 
and  has  21  teeth.  The  bevel  wheels  E  and  B  are  also  alike;  each  is  16.87 
inches  in  diameter  and  has  53  teeth.  The  drums  are  18  inches,  the  sheaves 
54  inches,  and  the  sheave  shafts  are  3^  inches  in  diameter.  The  lift  span 
weighs  68,000  pounds  and  each  counterweight  weighs  one-half  that  amount. 
Thus  the  span  would  be  perfectly  balanced,  if  the  mechanism  were  frictionless 
and  the  cables  without  stiffness  and  weight,  and  no  effort  would  be  required 
to  operate  the  bridge. 


Drurriia 

1-^ 


Ouf/m_e_  ofUft^  ^PPJl^ 


[A 
Drum%\ 


Drum 


Drum 


Diagram  of  Operating  Machinery. 


Tower 

Brat 


Sheave^ 

Counter- 
weight- 


■5 4-" Sheave 
Court  terweiqht 


7777777rr777777mTmmr, 

Side      Elevation. 
Fig.  351 


Cross    Section. 
Fig.  352 


In  the  following  computation  the  weight  of  cables  is  neglected.  Then  the 
tension  in  each  counterweight  cable  on  the  counterweight  side  of  the  sheaves 
is  one-fourth  of  34,000  pounds  or  8500  pounds.  When  the  span  is  being 
lifted,  the  tension  on  the  other  or  following  side  of  the  sheave  is  less  than  8500 
pounds.     Call  that  tension  T\   then 

8500  =  KT,  or  r  =  8500  -^  K 


222  •  Chap,  x 

(see  under  "  pulley"  above).  We  will  take  K  ==  1.06;  then  T  =  8020  pounds, 
and  hence  the  hft  on  the  span  due  to  counterweights  =  8  X  8020  =  64,160 
pounds.  This  leaves  68,000  —  64,160  =  3840  pounds  to  be  furnished  by  the 
four  up-haul  cables,  or  960  pounds  apiece. 

Let  a  and  b  —  respectively  any  corresponding  displacements  of  the  effort 
at  the  hand  lever  and  the  resistance  960  pounds  at  each  drum;   then 

b         3.43         21         18 

Hence,  if  the  mechanism  were  frictionless  the  effort  Pq  required  to  produce 
960  pounds  tension  in  one  rope  =  960  H-  50  =  19.2  pounds;  and  the  effort  P 
required  to  produce  that  tension  by  means  of  the  actual  mechanism  =  19.2  -^  e, 
where  e  =  the  efi&ciency  of  that  part  of  the  mechanism  which  transmits  from 
P  to  the  resistance  960  pounds.  The  efficiency  of  each  pair  of  gears  and 
necessary  bearings  we  take  as  0.95;  the  efficiency  of  a  drum  about  i  -h  1.03  = 
0.97;  hence e  =  0.95  X  0.95  X  0.97  =  0.875.  Therefore  P  =  19.2  -^  0.875  == 
22  pounds,  and  the  effort  (at  the  lever)  required  to  develop  a  tension  of  960 
pounds  at  the  four  driuns  =  4  X  22  =  88  pounds. 

The  computation  can  also  be  made  as  follows:  —  We  regard  the  total  force 
Q  exerted  at  the  hand  lever  and  the  force  of  gravity  on  the  counterweights  as 
two  efforts  which  overcome  the  (useful)  resistance  (gravity  on  the  lift  span) 
and  the  wasteful  resistances  in  the  entire  mechanism.  For  any  rise  b  of  the 
lift-span  the  counterweights  descend  an  equal  distance  and  the  hand-lever 
effort  works  through  a  distance  50  b;  and  since  the  efficiency  of  a  sheave  = 
I  -^  1.06  =  0.944,  we  have 

QX  SobX  0.875  -f  2  X  34,000  XbX  0.944  -=  68,000  X  6,  or  (J  =  87  pounds. 

45.   Kinetic  Friction 

§  I.  Kinetic  Friction,  or  Friction  of  Motion,  is  the  friction  between  two 
bodies  when  sliding  actually  occurs.  The  coefficient  of  kinetic  friction  for  two 
bodies  is  the  ratio  of  the  kinetic  friction  to  the  corresponding  normal  pressure 
between  them.  The  angle  of  kinetic  friction  is  the  angle  between  the  normal 
pressure  and  the  total  pressure  (resultant  of  the  normal  pressure  and  the 
kinetic  friction).  One  of  the  so-called  laws  of  friction  states  that  the  kinetic 
coefficient  is  less  than  the  static  coefficient  (Art.  19),  and  implies  that  there  is 
a  sudden  or  abrupt  change  in  the  values  of  the  coefficients.  Experiments  by 
Jenkin  and  Ewing*  on  the  kinetic  coefficients  at  speeds  as  low  as  0.0002  foot 
per  second  (about  f  foot  per  hour)  lead  them  to  conclude  that  "it  is  highly 
probable  that  the  kinetic  coefficient  gradually  increases  when  the  velocity 
becomes  extremely  small,  so  as  to  pass  without  discontinuity  into  the  static 
coefficient."     Experiments  by  Kimballf  also  indicate  that  there  is  no  abrupt 

*  Phil.  Trans.  Roy.  Soc,  1877,  Vol.  167,  Part  2. 
t  Am.  Jour.  Set.,  1877,  Vol.  13,  p.  353. 


Art.  45 


223 


change  from  static  to  kinetic  coefficient.  Moreover,  they  show  that  the 
kinetic  coefficient  may  be  greater  than  the  static.  Galton  and  Westinghouse 
experiments*  indicate  that  the  coefficient  for  dry  surfaces  probably  decreases 
progressively  from  the  value  of  the  static  coefficient  as  the  velocity  increases. 
See  the  following  table  of 

Coefficients  of  Friction  at  Various  Speeds 
Cast-iron  Brake  Shoes  on  Steel-tired  Wheels 


Velocity 

Coefficients 

Number  of 

tests 

Miles  per  hour 

Feet  per  second             Max 

mum 

Minimum 

Mean 

o-f- 

0+ 

0.330 

10 

14-5                    0 

281 

0. 161 

.242 

54 

20 

29 

240 

■133 

.192 

69 

30 

44 

196 

.098 

.164 

94 

40 

59 

194 

.088 

.140 

70 

50 

73 

153 

.050 

.116 

55 

60 

88 

123 

.058 

.074 

12 

The  foregoing  coefficients  are  based  on  observations  taken  very  soon  after 
application  of  the  brakes.  The  wide  variation  from  minimimi  to  maximum 
value  at  any  velocity  was  due  in  part  to  the  different  intensities  of  pressure 
employed  at  that  velocity;  in  general  the  coefficient  decreases  with  increase 
in  intensity  until  seizing  occurs. 

Continual  rubbing  of  dry  surfaces  abraids  them  and  decreases  the  coefficient 
of  friction.  This  effect  is  clearly  shown  in  the  following  table,  also  taken  from 
the  Galton  and  Westinghouse  experiments. 

Coefficient  of  Friction  as  Affected  by  Time  of  Rubbing 
Cast-iron  Brake  Shoes  on  Steel-tired  Wheels 


Miles  per  hour 

Time  after  applying  brakes 

0+ 

5  seconds 

10  seconds 

15  seconds 

20  seconds 

20 
27 
37 
47 

60 

0.182 
.171 

•152 
.132 
.072 

0.152 
.130 
.096 
.080 
.063 

0133 
.119 
,083 
.070 
.058 

O.I16 
.081 
.069 

0.099 
.072 

The  discrepancies  between  the  two  foregoing  tables  are  due  in  part  to  the 
fact  that  the  values  for  time  o  +  in  the  second  table  are  based  on  compara- 
tively few  experiments. 

The  following  table  gives  coefficients  for  several  different  materials;    it 
shows  also  influence  of  intensity  of  pressure,  velocity  and  water  lubrication. 


Proc.  Inst.  Mcch.  Engrs.,  1S79,  p.  170. 


2  24 


Chap,  x 


Coefficients  of  Friction  * 
Various  Brake-shoe  Materials  on  Steel-tired  Wheels 


Materials 


Cast  iron. 
Cast  iron. 

Oak 

Oak 

Poplar. . . . 
Poplar . . . . 
Cast  iron. 
Cast  iron. 

Oak 

Oak 

Poplar. . . . 
Poplar. . . . 


Pressure, 
pounds  per 
square  inch 


lO 

40 

10 

40 

10 

40 

20 

80 
40 

120 

40 

120 


Velocity,  miles  per  hour 


43 
36 
60 

43 


32 

30 

037 

073 

041 

070 


15 


0.37 
•30 
55 

40 

72 

53 
28 
26 
032 

055 
038 

053 


Lubrication 


none 

none 

none 

none 

none 

none 

water 

water 

water 

water 

water 

water 


*  From  Experiments  by  Ernest  Wilson,  Engr.  News,  1909,  Vol.  62,  p.  736. 

Coefficients  of  Kinetic  Friction  (Rough  Averages) 
Compiled  by  Rankine  from  Experiments  by  Morin  and  others 


Wood  on  wood,  dry  .  . 

soapy 
Metals  on  oak,  dry  . . . 

wet .  .  . 

soapy . 
Metals  on  elm,  dry. . . 
Hemp  on  oak,  dry. . . . 

wet. . . . 


0 

25-0.50 

2 

5-     -6 

24-      . 26 

2 

2  -  .25 

53 

33 

Leather  on  oak 

Leather  on  metals,  dry. .  . 

wet  .  . 

greasy 

oily .  .  . 
Metals  on  metals,  dry. . .  . 

wet .  .  .  . 


).  27-0.38 
■56 
•36 
■23 
■15 
.15-0.2 

•3? 


The  following  table  gives  coefficients  for  eight  ship  launchings  and  hence 
for  similar  cases.     The  lubricant  used  seems  to  have  been  mainly  tallow. 

Coefficients  from  Launching  Data.* 


Load,  tons  per 
square  foot 

Coefficient  of 
friction 

Load,  tons  per 
square  foot 

Coefficient  of 
friction 

0.50 

0.79 
1.03 
1.29 

0.0532 
.0487 
.0400 
.0407 

1.62 
2.05 
3.56 
4  50 

0.0370 
.0324 
.0257 
.0217 

*  From  Peabody's  Naval  Architecture. 

§  2.  Pivot  and  Journal  Friction.  —  Pivots.  —  Let  W  =  load,  ju  =  coeffi- 
cient of  friction,  (i)  In  the  case  of  a  flat  pivot  (Fig.  353)  the  average  pressure 
per  unit  area  of  contact  is  W/tR^.  On  any  element  of  area  dA  the  normal 
pressure  =  (W/irR^)  dA  (supposing  that  the  total  pressure  is  uniformly  dis- 
tributed), and  the  frictional  resistance  on  the  element  =  iJ.(W/irR^)dA.    The 


Art.  45 


225 


moment  of  this  resistance  about  the  axis  of  the  shaft  =  iJ.(W/irR^)dA-p. 
take  dA  =  pdd-dp;  then  the  total  resisting  moment  = 


We 


r  Tm 

Jo       Jo 


w 

-iddp'^dp 


pW-R. 
3 


Thus  the  actual  resistance  may  be  regarded  as  a  single  force  =  pW  with  an 
arm  —  ^R;  and,  for  example,  the  work  done  against  friction  per  revolution 
or  the  power  lost  may  be  computed  simply  on  that  basis.  Thus  the  work  done 
per  revolution  =  |  irpWR,  and  the  power  lost  =  |  irpWRn  where  n  =  number 
of  revolutions  per  unit  time. 

(ii)   In  a  similar  way  we  might  determine  the  resisting  (frictional)  moment 
in  a  collar  bearing  pivot  (Fig.  354).     We  would  find  the  moment  to  be 

1pW{R^-r^)-^  {R^-r^). 

Hence  we  may  regard  the  resistance  as  a  single  force  =  pW  with  an  arm 

f  {R^  -  r')  ^  (i?2  _  r'~). 


Fig.  353  Fig.  354 


Fig.  355         Fig.  356 


(iii)  In  the  conical  pivot  (Fig.  355),  the  total  normal  pressure,  and  hence  the 
friction  too,  is  increased  by  wedge  action.  Let  p  =  the  intensity  of  normal 
pressure  at  any  point  of  the  contact,  regarded  as  constant.  Then  the  normal 
pressure  on  an  elementary  area  dA  =  pdA .  Since  the  friction  has  no  vertical 
component,  the  vertical  component  of  the  normal  pressures  on  all  the  elemen- 
tary areas  =  W;   that  is, 

W 

pdA  •  sina  =  W  =  pA  sin  a,  or  p  = 


P 


A  sin  a 

But  A  sin  a  =  the  horizontal  projection  of  the  actual  surface  of  contact. 
Hence  the  intensity  of  the  normal  pressure  is  independent  of  a,  the  pivot 
angle.  For  Fig.  355,  p  =  W/tR-;  hence  the  normal  pressure  on  the  elemen- 
tary area  dA  is  {W/irR-)dA  and  the  frictional  resistance  =  p(W/TrR^)dA. 
The  moment  of  this  resistance  about  the  axis  of  the  shaft  =  p{W/TR^)dA-p, 
and  the  entire  resisting  moment  =  the  integral  of  this  expression.  For  sim- 
plicity in  integration,  imagine  dA  to  be  of  such  shape  that  its  horizontal 
projection  equals  pdd-dp  (see  Fig.  355).  Then  sin  a-dA  =  pdd-dp,  and  the 
resisting  moment  = 

^^pWddp^dp^  pW 
irR^  sin  a        sin  a  3 


Jr»2  7r      pL 
0       Jo 


'-R. 


226 


Chap,  x 


Hence  we  may  regard  the  resistance  as  a  single  force  p.W/sm  a  with  an  arm 
^  R 

(iv)  In  a  similar  way  we  may  compute  the  resisting  (frictional)  moment 
in  the  case  of  a  frustrated  conical  pivot  (Fig.  356).  We  would  find  that  the 
resisting  moment  = 

IjW  2  R^  —  r^ 
sin  a  3  i?"  —  r^ 

Hence  we  may  regard  the  friction  as  a  single  force  =  nW/sin  a  with  an  arm 
I  (R^  -  f^)/{R^  -  r^). 

Journal  Friction.  —  We  do  not  attempt  to  compute  the  normal  pressure  and 
frictional  resistance  at  each  point  of  a  journal  bearing  and  then  the  resisting 
moment  as  in  the  case  of  pivots.  So-called  coefficients  of  journal  friction  have 
been  determined  from  direct  experiments  on  journal  friction.  This  coefficient 
is  the  ratio  of  the  frictional  resistance  to  the  pressure  between  journal  and  the 
bearing.  Thus  in  a  certain  experiment  there  were  20  babbitt  bearings  sus- 
taining a  2yV-inch  shaft;  the  load  per  bearing  was  2,000  pounds,  and  it  was  found 
that  1 1 24  watts  were  required  to  run  the  shaft  at  350  revolutions  per  minute. 
All  the  power  was  used  to  overcome  the  journal  friction.  Since  11 24  watts  = 
49,600  foot-pounds  per  minute  and  350  revolutions  per  minute  corresponds  to 
a  (shaft)  surface  velocity  of  223  feet  per  minute,  the  total  frictional  resistance 
=  49,600 -T-  223  =  222  pounds  or  I  I.I  pounds  per  bearing.  Hence  the  coefficient 
of  journal  friction  in  this  particular  instance  was  ii.i  ^  2000  =  0.0055. 

The  pressure  between  a  journal  and  its  bearing  is  not  uniformly  distributed 
over  the  surface  of  contact.  By  nominal  intensity  of  pressure  ("pressure" 
for  brevity)  is  meant  the  whole  pressure  divided  by  the  product  of  the  length 
and  diameter  of  the  bearing.  Thus  in  the  experiment  just  mentioned,  the 
length  of  each  bearing  was  9!^  inches;    hence  the  nominal  intensity  was 


2000 


(2tV  X  9§i)  =  90  pounds  per  square  inch. 


It  has  been  found  from  numerous  experiments  that  coefiicients  of  journal 
friction  depend  on  (i)  the  method  of  lubrication,  (ii)  the  lubricant,  (iii)  its 
temperature,  (iv)  the  velocity  of  rubbing,  and  (v)  intensity  of  pressure  on  the 

bearing. 

(i)  Tower*  and  Goodmanf  report  the  following  relative  coefficients  as 
showing  effect  of  the  method  of  applying  the  lubricant: 


Method 


Bath 

Saturated  pad 
Ordinary  pad . 
Siphon 


Tower 


I  .00 

6.48 
7.06 


Goodman 


I  .00 
1.32 
2.21 
4.20 


*  Proc.  Inst.  Mech.  Engrs.,  1883. 

t  Mechanics  Applied  to  Engineering,  1896. 


Art.  45 


227 


(ii)   The  following  table  (according  to  Tower)  indicates  how  the  coefficient 
depends  on  the  lubricant.     Numbers  are  relative. 


sperm  oil 

Rape  oil 

Mineral  oil 

Lard  oil 

Mineral  grease 

1. 00 

I  .06 

1.29 

1-35 

2.17 

The  journal  was  steel;  gun  metal  brass  embracing  somewhat  less  than  one- 
half  the  circimiference  of  the  journal;  speed  300  revolutions  per  minute;  nom- 
inal loads  from  100  to  310  pounds  per  square  inch;  oil  temperature  90°  F.; 
bath  lubrication.  Tower  states  also:  "the  numbers  represent  the  relative 
thickness  or  body  of  the  various  oils,  and  also  in  their  order,  though  perhaps  not 
in  their  numerical  proportions,  their  relative  weight-carrying  power.  Thus 
sperm  oil,  wliich  has  the  highest  lubricating  power,  has  the  least  weight-carry- 
ing power;  and  though  the  best  oil  for  light  loads  would  be  inferior  to  the  thicker 
oils  if  heavy  pressures  or  high  temperatures  were  to  be  encountered,  (iii) 
The  coefficient  decreases  with  increased  temperature  (see  Figs.  357  and  359). 
But  if  the  temperature  gets  so  high  as  to  lower  the  viscosity  greatly,  then  the 
lubricant  gets  squeezed  out  and  the  coefficient  increases,  (iv)  In  general  the 
coefficient  increases  with  increase  of  speed  (see  Figs.  358  and  360  and  accom- 
panying explanations).  But  at  the  lower  speeds  the  coefficient  may  decrease 
with  increase  of  velocity  (see  Fig.  358).  (v)  The  coefficient  decreases  with 
increase  of  intensity  of  pressure  (see  Figs.  358  and  361).  But  the  intensity 
may  become  so  great  that  the  lubricant  gets  squeezed  out,  and  then  the  co- 
efficient increases  and  seizing  occurs. 


50        75         100        125        150 

Degrees  Fahrenheit. 
All  ValoclHes,  846  fr.permin. 

Fig.  357 


\ 

in-^ 

■ 

\ 

|4.^JS 

5;^ 

l\ 

^ 

Z 

Li-- 

■ 

\y 

<^ 

-^ 

bO 

lA-Z 

is^ 

-^ 

356 

)         100        200        300       400       500       600        700       80O 

Feet  per  Minute. 
All  Temperatures,  77° F. 

Fig.  3s8 


Figs.  357  and  358  are  after  Stribeck.*  Gas-motor  oil  and  ring  oiler  were 
used  in  the  experiments.  Figs.  359,  360,  and  361  are  after  Lasche.f  They 
show  respectively  how  the  coefficient  depends  on  temperature,  velocity,  and 

*  Z..  dV .  d.  I.,  1902,  Vol.  46,  p.  1341. 
\  Z.dV.  d.  /.,  1902,  Vol.  46,  p.  1881. 


227a 


Chap,  x 


pressure.     The  lubrication  was  forced;    journal  and  bearing  combinations  as 
follows: 


Number 
Journal . 
Bearing. 


I 

steel 
white  metal 


II 

nickel  steel 
white  metal 


III 

nickel  steel 


IV 

nickel  steel 
bronze 


V 

wrought  iron 
white  metal 


The  heavy  line  in  each  figure  represents  the  average  law  for  the  five  combina- 
tions, and  the  other  two  curves  relate  to  the  two  combinations  departing  most 
from  the  average  result. 


0.015 


^  g  0.010 


it  a.  0.005 


\ 

V 

^- 

'^^, 

^*>._ 
*«^ 

0.015 


0.010 


0.005 


y 

^''' 

„»-" 

"H 

0.03 


50     75      100      150     175 

Temperature,  Degrees  F. 

All  Pre55ur3s,  92.  Slbs.persq.in. 
All  Velocities,  I97ftpermin. 

Fig.  359 


)      1000     2000    3000    4000    5000 

Velocity,  ft.permin. 

All  Pressures,  9Z.5lbs.persq.  in. 
All  Temperatures,  112° F. 

Fig.  360 


0        50      100      150 

Pressure,  Ibs.per  sq.in . 

All  Temperatures,  II Z" F. 
All  Velocities,  197 ft.permin. 

Fig.  361 


Roller  and  Ball  Bearings  generally  develop  less  resistance  to  turning  than 
ordinary  bearings.  For  descriptions  of  roller  and  ball  bearings  and  information 
on  their  coefficients  of  resistance,  students  are  referred  to  works  on  machine 
design;  see  also  Kent's  Pocket  Book,  or  the  American  Civil  Engineer's  Pocket 
Book  for  coefl&cients  of  resistance.  But  to  furnish  some  notion  of  relative 
values  the  two  following  tables  are  given.  In  the  experiments  from  which  the 
first  was  compiled*  the  speed  was  560  revolutions  per  minute,  and  the  loads 
were  from  113  to  456  pounds.  In  the  ball  bearingsf  the  balls  were  |  inch  in 
diameter  and  they  ran  in  grooves,  cross  sections  of  which  were  arcs  of  circles 
whose  radius  was  equal  to  two-thirds  the  diameter  of  the  balls;  the  circle  of 
the  centers  of  the  balls  was  4  inches  in  diameter. 

Coefficients  of  Journal  Friction 
Different  Bearings  Ball  Bearings 


Diameter 
of  journal 

.   Flexible 
rollers 

Solid 
rollers 

Babbitt 
bearing 

2if 

0.018 
.014 
.032 
.025 

0.022 

.021 
.027 

0.043 
.082 
.096 
.107 

Load  in 
pounds 

Revolutions  per  minute 

6S 

38s 

780 

840 

2,420 

4.500 

10,800 

0.0033 
.0017 
•  0015 

0.0035 
.0018 
.0015 

0.0037 
.0019 
.0015 
.0011 

*  Benjamin,  Machinery,  N.  Y.,  1905,  Vol.  12,  p.  62. 
t  Stribeck,  Z.  d  V.  d.  I.,  igoi,  Vol.  45,  p.  121. 


Art.  45 


22711^ 


Tests  to  determine  the  coefficients  of  friction  for  ball,  flexible  roller,  and 
babbitt  bearings  for  line  shafts  have  been  made  at  the  University  of  Wiscon- 
sin. The  diameter  of  the  shaft  was  2^^  inches,  the  speed  150  to  450  revolu- 
tions per  minute,  the  load  700  to  2250  pounds  per  bearing,  (33  to  100  pounds 
per  square  inch  for  the  babbitt  bearings);  the  extreme  (natural)  variation 
of  the  temperature  of  the  lubricants  was  from  65  to  no  degrees  Fahrenheit. 
For  absolute  values  of  the  coefficients  for  the  various  conditions  named,  see 
report  of  the  tests.*  The  relative  value  of  the  coefficients  for  four  condi- 
tions are  given  in  the  following  table : 


RELATIVE  VALUES   OF  COEFFICIENTS   OF  JOURNAL  FRICTION,   AT 
LOAD   OF   ABOUT    1200   POUNDS   PER   BEARING. 


Peripheral  Speed 

Lubricant  Temperature. 


BaH  bearings. 
F]exible  roller. 
Babbitt 


100  ft/min. 


77  deg. 


100  deg. 


I 

2-5 

3-6 


300  ft/min. 


77  deg. 


I 

2.7 
4  5 


100  deg. 


I 
3 

4 


*  Thomas,  Maurer,  and  Kelso,  Jour.  Am.  Soc.  Mech.  Engrs.  for  March,  1914. 


CHAPTER   XI 

MOMENTUM   AND   IMPULSE* 
46.   Linear  Momentum  and  Impulse 

§  I.  (Linear)  Momentum.  —  By  momentum  of  a  moving  particle  is  meant 
the  product  of  its  mass  and  velocity.  We  regard  momentum  as  having  direc- 
tion, namely,  that  of  velocity;  thus,  momentum  is  a  vector  quantity.  By 
momentum  of  a  collection  of  particles  is  meant  the  vector-sum  of  the  momen- 

tums  of  the  particles.  For  example,  let  m' 
and  m"  =  the  masses  of  two  particles  (Fig. 
362),  v'  and  v"  =  the  velocities  of  the  par- 
ticles at  a  certain  instant,  and  suppose  that 
AB  =  m'v'  and  BC  =  m"v"  according  to  some 
convenient  scale;  then'  AC  represents  the 
^  momentum  of  the  two  particles. 

Since  the  component  of  the  vector  AC  along  any  line  equals  the  algebraic 
sum  of  the  components  of  the  vectors  AB  and  BC  along  that  line,  it  follows 
that  the  component  of  the  momentum  of  a  pair  of  particles  along  any  line 
equals  the  algebraic  sum  of  the  components  of  their  momentums  along  that 
line.  Obviously,  this  proposition  can  be  extended  to  a  collection  of  any  number 
of  particles.  A  simple  expression  for  this  component  can  be  arrived  at  as 
follows:  Let  m',  m" ,  m'  ",  etc.  =  the  masses  of  the  particles;  v' ,  v",  etc.  =  their 
velocities;  and  v'^,  -a" x,  etc.  =  the  components  of  these  velocities  along  any  Une 
X.  Then  the  component  of  the  momentum  of  the  collection  along  this  line  = 
^'^'^  _|_  yn"v"x  +  .  .  .  Now  if  x' ,  x",  etc.  =  the  x  coordinates  of  the 
moving  particles,  and  x  =  the  x  coordinate  of  the  mass-center,  all  at  the  same 
instant,  then  m'x'  +  m"x"  +  .  .  .  =  xHm  (Art.  34);  and  differentiating 
with  respect  to  /,  we  get  m'dx'/dt  -\-  ni"dx"/dt  -|-      .     .     .     =  {dx/dt)llm,  or 

m'v'^  +  m"v"x  +....=  VjJZm  =  Mvx, 

where  M  =  2m  =  the  mass  of  the  collection.  That  is,  the  x  component  of  the 
momentum  of  the  collection  of  particles  equals  the  product  of  the  mass  of  the 
collection  and  the  x  component  of  the  velocity  of  the  mass-center.  Hence, 
the  component  momentum  is  just'  the  same  as  though  all  the  material  of  the 
body  were  concentrated  at  the  mass-center. 

In  the  case  of  a  body  having  a  motion  of  translation,  all  the  particles  have 
at  any  instant  velocities  which  are  equal  in  magnitude  and  the  same  in  direc- 
tion (Art.  35).     Hence  the  momentums  of  the  particles  are  parallel,  and  their 

*  This  chapter  is  not  prerequisite  to  Chapter  XII. 

228 


Art.  46  229 

vector  sum  =  m'v  +  m"v  +  .  ,  .  =  vTtm  =  Mv,  where  v  =  their  common 
velocity  and  M  =  the  mass  of  the  body. 

The  definition  of  momentum  impUes  that  the  unit  of  momentum  equals  the 
momentum  of  a  body  of  unit  mass  moving  with  unit  velocity.  The  magnitude 
of  the  unit,  therefore,  depends  on  the  units  of  mass  and  velocity  used.  No 
single  word  has  been  generally  accepted  for  any  unit  of  momentum.  The 
dimensional  formula  for  momentum  is  F'T'  (see  appendix  A),  that  is,  a  unit 
momentmn  is  one  dimension  in  force  and  one  in  time.  Hence,  any  unit  of 
momentum  may  be  and  commonly  is  called  by  names  of  the  units  of  force  and 
time  used.  Thus  the  unit  of  momentum  in  the  C.G.S.  system  is  called  the 
dyne-second;   in  the  ''engineers'  system,"  the  pound  (force)  -second,  etc. 

In  Art.  34  it  is  explained  that  the  acceleration  of  the  mass-center  of  any 
collection  of  particles  does  not  depend  at  all  on  the  forces  which  the  particles 
exert  upon  each  other  but  on  the  external  forces;  also  that  the  algebraic  sum 
of  the  components  of  the  external  forces  along  any  line  equals  the  product  of 
the  mass  of  the  system  and  the  component  of  the  acceleration  of  the  mass- 
center  along  that  line,  that  is, 

n  +  i^"x+     .     .     .  =  Mi., (i) 

where  F'x,  P"x,  etc.,  are  the  components  of  the  external  forces  along  a  line  x, 
and  ax  is  the  x  component  of  the  acceleration  of  the  mass-center.  Now  ax 
equals  the  rate  at  which  the  x  component  of  the  velocity  of  the  mass-center 
changes,  that  is,  ax  —  dvx/dt,  where  Vx  is  the  x  component  of  the  velocity  of  the 
mass-center;  hence,  Max  =  M dvx/dt  =  d{Mvx)/dt;  and  finally 

F'x  +  F"x  +     .     .     .     =  d{,Mvx)/dt (2) 

But  Mvx  is  the  x  component  of  the  momentum  of  the  system,  and  d{Mv^/dt 
is  the  rate  at  which  that  component  changes;  hence  the  algebraic  sum  of  the 
components  of  the  external  forces  along  any  line  x  equals  the  rate  at  which  the  x 
component  of  the  linear  momentum  changes. 

The  principle  just  arrived  at  (equation  2)  was  derived  from  the  law  of  motion 
of  the  mass-center  (equation  i),  and  it  is  essentially  an  alternative  form  of  the 
law.  But  practically  the  former  seems  to  apply  more  simply  in  certain  cases 
as  the  following  examples  show. 

Fig.  363  represents  a  jet  of  water  impinging  against  a  flat  plate.  Required 
the  pressure  of  the  jet  upon  the  plate.  Let  W  =  the  weight  of  water  impinging 
per  unit  time,  v  =  the  velocity  of  the  water  in  the  jet,  and  a  =  the  angle 
between  the  jet  and  the  plate  as  indicated.  We  suppose  that  the  water  does 
not  rebound  from  the  plate  with  any  considerable  velocity;  then  the  momen- 
tum of  the  water  after  striking  has  no  component  normal  to  the  plate.  The 
momentum  of  an  amount  of  water  equal  to  W  before  striking  is  {W/g)v,  and 
the  component  of  that  momentum  along  the  normal  to  the  plate  =  (W/g)v 
sin  a;  hence  the  change  in  the  (normal)  component  momentum  is  iW/g)v  sin  a. 
This  change   takes  place  in  unit  time;    therefore,  it  is   the  rate  at   which 


230 


Chap,  xi 


momentum  along  the  normal  is  changed,  and  also  the  value  of  the  normal 
pressure  of  the  plate  against  the  jet.  The  jet  exerts  an  equal  (normal)  pressure 
against  the  plate.  If  the  plate  is  rough,  then  the  water  also  exerts  a  frictional 
force  on  the  plate. 


-w/m//w^/w7i^;/w,. 


Fig.  363 


For  another  example,  we  will  determine  the  pressure  on  a  bend  in  a  pipe 
by  water  flowing  through  it  at  constant  velocity.  Let  W  =  the  weight  of 
the  water  flowing  past  any  section  of  the  pipe  per  unit  time;  v  =  velocity 
of  the  water,  assumed  to  be  the  same  at  all  points  of  inlet  and  outlet  cross 
sections  of  the  bend;  and  a  =  the  angle  of  the  bend  (Fig.  364).  Also  let 
A^  =  the  time  required  for  the  body  of  water  AB  to  move  into  the  position 
A'B'.  The  momentum  of  the  body  of  water  at  the  beginning  of  the 
interval  =  that  of  AA'  -\-  that  of  A'B;  its  momentum  at  the  end  of  the 
interval  =  that  oi  A'B  -\-  that  of  BB' .  Hence  the  change  in  the  momentum 
of  the  body  of  water  in  the  time  Ai  =  momentum  of  BB'  —  momentum  of 
AA'.  These  momentums  respectively  are  in  the  direction  BB'  and  AA';  each 
equals  {WM/g)v.  Hence  the  change  of  momentum  under  consideration  is 
represented  by  the  vector  MN  where  OM  and  ON  represent  the  two  mo- 
mentums just  mentioned.  But  MN  =  2{0M)  sm\a;  hence  the  change  = 
2{WM/g)v  sin  ^  a,  and  the  rate  at  which  the  change  occurs  =  2{W/g)v  sin  ^  a. 
The  direction  of  this  rate  is  MN;  it  bisects  the  angle  a.  This  rate  of  change 
of  momentum  is  maintained  by  the  forces  acting  on  the  body  of  water  in  A  'B. 
Those  forces  consist  of  gravity  G,  the  pressures  Pi  and  P2  (of  the  water)  on 
the  front  and  rear  faces  of  the  body,  and  the  pressure  P  of  the  bend  upon  it. 
Their  resultant  R  =  2{W/g)v  sin  |  a,  and  R  bisects  a.  If  R,  G,  Pi  and  P2  are 
known  then  P  can  be  determined.  For  it  is  such  a  force  which  compounded 
with  G,  Pi  and  Pi  gives  R.     The  pressure  of  the  water  on  the  bend  =  —  P. 

For  another  example,  we  take  the  jet  propeller  of  a  ship.  This  consists 
essentially  of  a  pump  which  takes  in  water  from  the  sea  and  ejects  it  from 
nozzles  toward  the  rear  (to  propel  the  ship  forward).  Let  W  =  weight  of 
water  so  ejected  per  unit  time,  v  =  velocity  of  the  ship,  and  V  —  velocity  of 
the  ejected  water  relative  to  the  ship.  The  absolute  velocity  of  the  jet  (rela- 
tive to  the  sea)  =  V  —  v.  Hence  the  amount  of  momentum  produced  by  the 
pimiping  plant  (pump,  pipes,  etc.)  per  unit  time  =  (W/g)  (V  —  v).  The 
direction  of  this  is  horizontal  and  backward;  hence  the  plant  exerts  a  force  on 
the  body  of  water  within  the  passages  at  any  instant  equal  to  (W/g)  {V  —  v); 
the  water  exerts  an  equal  force  forward  on  the  passages. 


Art.  46  231 

If  the  algebraic  sum  of  the  components  —  along  any  line  —  of  the  external 
forces  acting  on  a  body  equals  zero,  then  the  rate  of  change  of  the  component 
momentum  (along  that  Une)  equals  zero;  hence,  if  the  sum  remains  zero  for 
any  interval  of  time,  the  component  momentum  remains  constant.  This  is 
known  as  the  principle  of  conservation  of  linear  momentum.  It  follows  that 
if  there  are  no  external  forces  acting  on  the  body,  its  linear  momentum  remains 
constant.  The  grand  illustration  of  this  principle  is  furnished  by  the  solar 
system.  Even  the  nearest  stars  exert  no  appreciable  attractions  on  the  solar 
system,  and  so  the  members  of  the  system  move  under  the  action  of  their 
mutual  attractions  only.  Accordingly,  the  component  of  the  momentum  of 
the  system  along  any  line  does  not  change;  the  linear  momentum  is,  therefore, 
constant  in  amount  and  direction.  It  follows  that  the  mass-center  of  the 
system  moves  uniformly,  and  in  a  straight  line. 

§  2.  (Linear)  Impulse.  —  If  the  magnitude  and  direction  of  a  force  are 
constant  for  any  interval  of  time,  then  the  product  of  the  magnitude  of  the 
force  and  the  interval  is  called  the  impulse  of  the  force  for  that  interval.  If 
the  magnitude  varies,  then  the  impulse  for  any  interval  equals  the  sum  of  the 
impulses  for  all  the  elementary  periods  of  time  which  make  up  the  interval ; 
that  is  impulse  = 

lim  [F'At  +  F"At  +  .     .     .     .     ]  =  j Fdt, 

where  F  =  the  varying  force.  If  the  direction  of  the  force  varies,  we  regard 
the  impulse  for  any  elementary  portion  of  time  as  a  vector  quantity  having 
the  direction  of  the  force,  and  then  in  principle  we  add  (vectorially)  the  ele- 
mentary impulses  for  all  the  portions  of  time  which  make  up  the  interval. 
That  is  to  say,  we  integrate  Fdt  vectorially,  arriving  at  a  definite  vector 
quantity. 

Units  of  impulse  depend  on  the  units  of  force  and  time  used.*     There  is  no 
current  single- word  name  for  any  unit  of  impulse.     Each 
unit  is  named  by  the  names  of  the  units  of  force  and  time 
involved  in  it.     Thus,  in  the   C.G.S.  system  the  unit  of     ; 
impulse  is  the  dyne-second;   in  the  "engineers'  system" 
the  unit  of  impulse  is  the  pound  (force)  -second.  Fig* "365 

It  is  evident  (Fig.  365)  that  the  elementary  inpulse  F  dt 
is  the  resultant  of  the  impulses  of  the  x  and  y  components  of  F  (or  .v,  y, 
and  2  components,  if  preferred).     Hence  the  x,  y,  and  z  components  respec- 
tively of  the  impulse  of  F  equal  the  impulses  of  the  components  of  the  force 
F.     If  we  integrate  equation  (2)  over  any  interval  ti  -  ti,  say,  we  get 

f'^F'^dt  +  rV".(f^  -f-     .     .     .     =  MvJ'  -  M^J  =  A  (M^x),      (3) 

where  vj  and  vj'  =  the  x  velocities  of  the  center  of  gravity  of  the  system  at 
times  ti  and  fe  respectively.     Equation  (3)  can  be  put  into   the  following 

*  See  appendix  A. 


232  Chap,  xi 

principle  of  {linear)  impulse  and  momentum.  The  algebraic  sum  of  the  com- 
ponents —  along  any  line  —  of  the  impulses  of  the  external  forces  acting  on 
any  system  of  particles  equals  the  increment  in  the  component  of  the  mo- 
mentum of  the  system  along  that  same  line,  the  sum  and  the  increment  re- 
ferring to  any  interval  of  time. 

The  principle  of  impulse  and  momentum  answers  such  questions  as,  —  how 
much  velocity  in  a  given  time?  or  how  much  time  to  produce  a  certain  ve- 
locity? For  example,  it  is  required  to  ascertain  how  much  time  is  required 
to  give  a  velocity  of  40  feet  per  second  to  a  certain  body  by  sliding  it  along  a 
horizontal  rail  by  means  of  a  constant  push  of  20  pounds,  the  body  weighing 
100  pounds  and  the  frictional  resistance  of  the  rail  being  8  pounds.  The  ex- 
ternal forces  acting  on  the  body  are  gravity,  the  push,  and  the  reaction  of  the 
rail,  the  horizontal  and  vertical  components  of  which  are  friction  and  the 
"  normal  pressure."  Only  the  impulses  of  the  push  and  friction  have  com- 
ponents along  the  line  of  motion;   hence 

20  /  —  8  /  =  (100/32.2)  40, 

where  /  =  the  required  time.  Therefore  /  =  10.3  seconds.  Solution  of  such 
a  problem  by  earlier  methods  of  this  book  would  be  as  follows:  Let  a  =  the 
acceleration;  then  a  =  (20  —  8)  -^  (100/32.2)  =  3.86  feet  per  second  per 
second.     Hence  /  =  40  -^  3.86  —  10.3  seconds. 

47.   Impact  or  Collision 

§  I.  Blow.  —  Momentum  of  a  blow,  energy  of  a  blow,  and  especially 
force  of  a  blow  are  terms  generally  used  more  or  less  vaguely.  But  when  one 
of  the  two  colliding  bodies  is  fixed,  then  the  first  two  terms  are  taken  to  mean 
the  momentum  and  the  kinetic  energy  respectively  of  the  moving  body  just 
before  the  impact,  perfectly  definite  quantities.  If  the  motion  is  one  of 
translation,  these  are  Mv  =  {W/g)v  and  \  Mv^  =  |  {W/g)v'  respectively.  If 
in  a  numerical  case  we  write  g  =  32.2  (feet  per  second  per  second),  v  should  be 
expressed  in  feet  per  second;  W  may  be  expressed  in  any  force  unit.  If  the 
pound  is  used,  then  the  momentum  is  in  pound-seconds  and  the  energy  in 
foot-pounds. 

Force  of  a  blow  means  the  pressure  which  two  colliding  bodies  exert  upon 
each  other.  The  pressure  changes  during  the  collision.  Analysis  of  this 
variation  is  beyond  the  scope  of  this  book.  We  will  deal  only  with  average 
values  of  the  force  of  a  blow.  In  the  first  place,  it  should  be  noted  that  there 
are  two  average  values  of  the  force  of  a  given  blow,  —  a  space-average  and  a 
time-average.  We  explain  the  distinction  by  means  of  an  example,  but  we 
choose  the  simpler  case  of  a  varying  horizontal  pull  dragging  a  body  along  a 
smooth  horizontal  surface  instead  of  a  blow.  Let  us  suppose  first  that  the 
pull  varies  uniformly  with  respect  to  time,  from  a  zero  value  to  40  pounds  in 
20  seconds  (see  Fig.  366).  Then  the  time-average  is  represented  by  the 
average  ordinate  to  the  line  which  shows  how  the  force  varies  with  respect  to 


Art.  47 


233 


the  time;  hence  it  is  20  pounds.  We  wish  to  find  now  how  the  force  varies 
with  respect  to  distance.  Let  P  =  the  value  of  the  pull  at  any  time  t  after 
starting;  then  the  law  of  force  is  P  =  2  ^.  Also  let  M  =  mass  of  the  body; 
a  and  v  respectively  =  the  acceleration  and  velocity  at  any  time  t,  and 
s  =  the  displacement  up  to  that  time.     Then 


a  =  -r7  =  ir7i,     V  =  Yji",     and     5  =  — r-^ 


t\ 


The  total  displacement  (^i)  in  the  20  seconds  =  (1/3  M)  8000.     It  follows 
from  the  last  equation  that 

t  =  (3  Ms)^\    hence  P  =  2  (3  Ms)^. 

This  equation  determines  the  graph  shown  in  Fig.  367,  from  which  it  is  ap- 
parent that  the  space-average  force  is  more  than  20  pounds,  or  the  time-aver- 


40  lbs 

— 

^^ 

■30 

/" 

•20 

/ 

■\oy 

0       5 

10 

15 

SOsec 

Fig. 

366 

•-1000 -e-3M 
Fig.  367 


age.     The  space-average  equals  the  area  under  the  curve  divided  by  the  base 
of  the  area.     The  area  is 

/  '' P  ds^   I  "2  isMs)^   ds  =  1.5  (3M)^5i*; 

hence  the  space-average  =  1.5  (3  AT)^  ^i^  =  30  pounds. 

It  will  be  observed  that  the  space-average  is  that  constant  force  whose 
work  equals  the  work  done  by  the  (real)  varying  force  (see  Art.  40).  Like- 
wise the  time-average  is  that  constant  force  whose  impulse  equals  the  impulse 
of  the  (real)  varying  force.  Hence  the  space-average  equals  the  quotient 
of  the  work  done  by  the  force  (equal  to  the  kinetic  energy  produced  by  the 
force)  and  the  distance  through  which  the  force  acted;  and  the  time-average 
equals  the  quotient  of  the  impulse  of  the  force  (equal  to  the  momentum 
produced  by  the  force)  and  the  duration  of  the  impulse. 

Crushers  or  crusher  gages  are  small  cylinders  of  copper  or  lead  —  one  inch 
diameter  and  one  inch  high  are  common  proportions  —  used  in  a  way  de- 
scribed presently  to  determine  the  energy  and  force  of  a  blow.  Fig.  3 68  shows 
several  energy-compression  curves  for  i\-  by  i|-inch  lead  cylinders.*  Curve 
A-B  IS  a  so-called  static  curve,  and  was  obtained  by  crushing  a  lead  cylinder 
in  an  ordinary  testing  machine  at  0.05  inches  per  minute.  The  amount  of 
compressing  force  and  the  amount  of  the  compression  were  observed  at  fre- 
quent stages  during  the  test,  and  from  these  observations  the  amount  of  work 

*  American  Machinist,  Vol.  i^,  Part  i,  p.  436  (1910). 


234 


Chap,  xi 


done  on  the  cylinder  up  to  each  stage  was  computed.  Amounts  of  compression 
and  corresponding  amounts  of  work  were  plotted  to  determine  the  curve. 
Curve  C  is  a  static  curve  but  for  a  higher  speed.  D  is  a  so-called  dynamic 
curve.  It  was  obtained  from  drop  or  impact  tests  in  which  each  crusher  was 
subjected  to  a  blow  from  a   "hammer"   dropped  upon  it.     The  hammer 


c 
o 


1- 
O- 

e 

o 
U 


A 

t 

". 



AU 

t 

-c 

_ 

— 

^^ 

X\'< 

.1 

^- 

1— ^ 

^ 

•D 

^^ 

,»— 

— 

\J 

^ 

.^ 

' 

-— ^ 

■^ 

.0 

/ 

,^ 

^ 

^^^ 

-^ 

.8 

/ 

/• 

.^ 

^ 

J 

/ 

^^ 

'^ 

^ 

4 

^ 

^>' 

.6 

l/t 

7^ 

.4 

T 

.2 

• 

10000 


20000  30000 

Energ-y,  inch-pounds. 

Fig.  368 


40000 


50000 


weighed  1330  pounds,  and  the  maximum  drop  used  was  38  inches.  For  each 
test  the  amount  of  compression  c  was  observed  and  the  amount  of  work  1330 
{h  +  c)  was  computed,  where  //-  =  height  of  drop  to  the  cylinder;  and  this 
compression  and  work  were  plotted  for  one  point  on  the  curve.  Curve  E-F 
is  a  dynamic  curve  obtained  from  tests  in  which  the  hammers  were  lighter  and 
the  drops  higher  than  for  D. 

Crushers  are  used  as  follows  to  determine  the  energy  of  a  blow,  as  of  a  steam 
hammer  for  example.  The  crusher  is  placed  on  the  anvil  and  subjected  to  the 
blow.  Then  the  amount  of  the  compression  of  the  crusher  is  measured,  and 
the  corresponding  energy  is  read  off  from  the  appropriate  compression-energy 
curve  (previously  determined  from  tests  on  crushers  like  the  one  used).  The 
space-average  force  of  such  a  blow  equals  the  quotient  of  the  energy  of  the  blow 
and  the  compression  unless  cjh  is  not  a  small  fraction ;  in  that  case  the  space- 
average  =  W  Qi  -\-  c)  -^  c. 

A  very  skillful  use  of  the  crusher  was  made  by  Lieut.  B.  W.  Dunn  to  deter- 
mine not  only  the  average  but  the  actual  value  of  the  force  of  a  blow  at  any 
instant  during  the  impact.*  He  devised  apparatus  which  made  a  photo- 
graphic record  (space-time  graph.  Art.  29)  of  the  motion  of  a  hammer  during 
the  impact.  From  that  graph  he  deduced  the  velocity-time,  and  from  this 
the  retardation-time  graph;  then  the  force  of  the  blow  F  at  any  instant  from 
/r  =  p[7  -|-  {W/g)  a,  where  W  =  weight  of  hammer  and  a  =  retardation  at  the 
instant.  The  order  of  measurements  involved  in  this  apparatus  is  indicated 
by  these  circumstances  of  one  test:   amount  of  compression  of  crusher  about 

*  Journal  of  the  Franklin  Institute,  Vol.  144,  p.  321  (1897). 


Art.  47  235 

^  inch  and  time  of  impact  about  xmiTT  second;  the  weight  of  hammer  was 
^T,  pounds  and  the  drop  15  inches.  For  the  copper  crushers  used  the  maxi- 
mum pressure  occurred  just  before  the  end  of  the  compression,  and  its  value 
was  slightly  less  than  twice  the  space-average. 

§  2.  Motion  after  Collision.  —  In  this  section  we  discuss  the  changes  of 
motion  of  one  or  both  colliding  bodies  due  to  the  collision  in  certain  compara- 
tively simple  cases.  In  most  cases  of  collision  the  pressures  which  the  colliding 
bodies  exert  on  each  other  are  enormous  compared  with  other  forces  acting 
on  the  bodies.  For  example,  the  space-average  pressure  between  two  billiard 
balls  colliding  with  velocity  of  8  feet  per  second  is  about  1300  pounds.  There- 
fore in  discussing  changes  of  motion  of  the  bodies  during  collision  we  may 
neglect  the  other  (ordinary)  forces  acting  on  the  bodies,  gravity  for  example; 
that  is  we  regard  the  two  bodies  jointly  as  under  the  action  of  no  external 
forces.  Hence,  according  to  the  principle  of  conservation  (Art.  46),  the 
momentum  of  the  two  bodies  jointly  is  not  changed  by  the  impact. 

If  the  centers  of  gravity  of  two  bodies  about  to  collide  are  moving  along  the 
same  straight  line,  then  the  collision  or  impact  is  called  direct;  if  otherwise, 
oblique.  If  the  pressures  which  two  colliding  bodies  exert  upon  each  other 
during  impact  are  directed  along  the  line  joining  their  centers  of  gravity,  then 
the  impact  is  called  central;  if  otherwise,  eccentric.  These  are  the  kinds  of 
impact  called  simple,  above. 

Direct  Central  Impact.  —  We  assume  that  the  bodies  have  motions  of  trans- 
lation before  impact.  Since  the  impact  is  supposed  to  be  central,  the  pressure 
(of  impact)  on  each  body  acts  through  the  center  of  gravity  of  that  body  and 
does  not  turn  it.  Hence  the  motion  of  each  body  after  collision  is  one  of 
translation.     Let  A  and  B  be  the  two  bodies, 

Ml  and  M2  =  their  masses, 
Ui  and  ih.  =  their  velocities  just  before  impact, 
and  Vi  and  V2  =  their  velocities  just  after  impact  respectively. 

We  regard  these  velocities  as  having  sign;  velocity  in  one  direction  (along  the 
line  of  motion)  being  positive,  and  that  in  the  other  being  negative.  Then  the 
momentum  of  the  two  bodies  before  impact  =  MiUi  -f-  M2M2,  and  after  impact 
it  =  MiVi  -\-  M2V2.  Since  the  momentums  before  and  after  impact  are  equal, 
we  have 

MiVi  +  M2V2  =  MiUx  +  7I/2M2.  (i) 

The  foregoing  expressions  are  correct  whether  A  and  B  are  moving  in  the  same 
or  opposite  directions  before  or  after  the  impact.  Thus,  if  both  are  moving 
toward  the  right  before  impact,  at  8  and  10  feet  per  second  say,  their  momen- 
tum is  8  ^1/1  -|-  10  7I/2;  but  if  A  is  moving  toward  the  right  and  B  toward  the 
left,  their  momentum  is  8  Mi  —  10  M2. 

It  has  been  learned  experimentally  that  when  two  spheres  A  and  B  collide 
directly  and  centrally  the  velocity  of  separation  is  always  less  than  and  opposite' 
to  the  velocity  of  approach,  and  the  ratio  of  these  two  velocities  seems  to 


236 


Chap,  xi 


depend  only  on  the  material  of  the  two  spheres.  The  ratio  of  the  velocity  of 
separation  to  that  of  approach  (signs  disregarded)  is  called  coefficient  of  restitu- 
tion; it  is  generally  denoted  by  e.  The  following  are  approximate  values  of 
e  for  a  few  materials, 

glass  II,  ivory  |,  steel  and  cork  |,  wood  about  \,  clay  and  putty  o. 
Now  the  velocity  of  approach  equals  Ui  —  ih  (or  W2  —  Ui),  —  the  first  with 
reference  to  A  (regarded  as  fixed)  and  the  second  with  reference  to  B  (re- 
garded as  fixed)  — ,   and  the  velocity  of  separation  is  Vi  —  Vo   (or  v^  —  Vi). 
Since  these  velocities  are  opposite  in  direction,  we  have 

—  {vi  —  v^/{ui  —  U2)  =  e,     or     —  {vi  —  V2)  =  e  (th  —  U2).  (2) 

Equations  (i)  and  (2)  solved  simultaneously  for  the  final  velocities  Vi  and 
V2  give 
v.  =  u^-(i+e)j^^j^^{u,-u.);v2  =  i^-{i  +  e)j^-^^^^  (3) 

If  one  of  the  colHding  bodies  is  fixed,  say  B,  then  1^  =  o,  and  M2  is  the  mass  of 
B  and  its  supports,  infinitely  great.     Thus  we  have  Vy  =  —  eui. 

Oblique  Central  Impact.  —  We  assume  as  before  that  the  bodies  A  and  B 
have  a  motion  of  translation  before  impact;  then  the  pressure  on  each  during 
the  impact  acts  through  the  center  of  gravity  and  produces  no  turning.  Let 
Ui  and  1/2  =  the  velocities  of  A  and  B  before  impact;   Vi  and  F2  their  velocities 

after  impact;  Ui  and  M2  =  the  components  of  Ui 
and  U2  along  the  fine  of  impact  pressure  (joining 
the  centers  of  gravity  of  A  and  B  when  in  con- 
tact); V]_  and  V2  =  the  components  of  Vi  and  F2 
^'  along  that  line;  and  Wi  and  W2  =  the  components 

Pjg     .  of  Z7i  and  U2  at  right  angles  to  that  line.     See 

Fig.  369  which  represents  one  of  several  possible 
ways  of  oblique  collision.  Since  the  impact  pressure  on  either  body  has  no 
component  transversely  to  the  line  of  pressure  XX,  the  component  of  the 
momentum  of  either  body  at  right  angles  to  XX  is  not  changed.  Hence 
the  transverse  component  of  the  velocity  of  either  body  is  not  changed  by 
the  impact.  The  longitudinal  components  are  changed  as  in  direct  impact, 
and  vi  and  V2  are  given  by  equations  (3).  The  final  velocities  Fi  and  F2, 
therefore,  are  determined,  Vi  by  its  components  Vi  and  Wi,  and  V2  by  its 
components  %  and  W2. 
Loss  of  Energy  in  Impact.  —  Let  L  =  the  loss  of  kinetic  energy;  then 

L  =  (i  M,U^'  +  i  M2U2')  -  (^  MiFi^  +  \  M2V2'). 
Now  f/i^  =  wi2  -f  wi2,  C/2'  =  ^2^  +  ■W2\  Vi"  =  Vi"  +  Wi",  and  Fg^  =  V2^  +  W2^; 

hence 

Z  =  i  MM'  -  V)  +  h  M2W  -  ^2^). 
Substituting  for  Vi  and  V2  their  values  from  equation  (3)  and  simplifying  we  get 

.       .  „,  M1M2  ,  s. 


Art.  48 


237 


For  perfectly  elastic  bodies  (e  =  i),  L  =  o.  For  other  bodies  (i  —  e-)  is  not 
zero  but  a  positive  quantity;  and  since  (th  —  ito)  is  not  zero,  L  is  always  a 
finite  positive  quantity.  That  is,  in  every  collision  of  bodies  not  perfectly 
elastic  there  is  loss  of  kinetic  energy.  If  the  bodies  are  without  elasticity 
(e  =  o),  the  loss  =  |  [{M,M2)/(Mi  +  M.)]  (ih  -  u^y. 

The  foregoing  is  essentially  Newton's  analysis  of  impact.  Several  more 
recent  analyses  have  been  made  independent  of  any  coefificient  of  restitution 
but  taking  into  account  the  vibrations  set  up  in  the  colliding  bodies.  On 
account  of  the  difficulties  of  the  problem  they  include  only  impact  of  spheres 
and  cylinders  end  on.  Explanation  of  these  analyses  fall  beyond  the  scope 
of  this  book.* 

48.   Angular  Momentum  and  Impulse 

§  I.  Angular  Momentum.  — The  linear  momentum  of  a  moving  particle 
is  a  vector  quantity,  as  explained  in  Art.  46 ;  the  magnitude  of  the  momentum 
is  mv  (where  m  =  mass  of  the  particle  and  v  =  its  velocity),  and  the  direction  is 
that  of  the  velocity.  We  go  farther  now  and  assign  position  to  the  mo- 
mentum and  to  the  momentum-vector.  The  position,  or  position-line,  of  the 
momentum  of  a  moving  particle  is  the  line  through  the  particle  in  the  direction 
of  the  velocity.  Thus  the  Hnear  momentum  of  a  particle  is  a  "  locaUzed  " 
vector  quantity,  —  like  a  concentrated  force,  which  has  magnitude,  direc- 
tion and  a  definite  position,  or  line  of  action  as  it  is  more  commonly  called. 

We  apply  the  term  moment  of  momentum  to  a  product  which  is  analogous  to 
the  product  which  we  call  moment  of  a  force  about  a  line.     Thus  the  moment 
of  momentum  of  a  moving  particle  about  a  line  (or 
angular  momentimi  as  it  is  also  called)  is  the  product 
of  the  component  of  the  momentum  perpendicular 
to  the  Une  --  the  other  component  being  parallel 
to  it  —  and  the  distance  from  the  line  to  the  per- 
pendicular  component.      (Compare    definition    of 
moment  of  a  force  about  a  line,  Art.   8.)      For 
example,  let  0  (Fig.  370)  be  the  position  of  the 
moving  particle  at  a  given  instant,  OC  the  direc-       ( 
tion  of  its  velocity,  and  OABC  a  parallelogram  Fig.  370 

whose    sides    are    parallel    and   perpendicular   to 

the  line  LL\  an  axis  of  moments.  (QQ  is  a  plane  perpendicular  to  LL' 
represented  to  make  the  figure  more  plain.)  Then  according  to  some  scale 
OC  represents  the  momentum  mv,  and  OA  and  OB  represent  components 
of  mv  perpendicular  ajid  parallel  to  LL'  respectively.  The  angular  momen- 
tum of  the  particle  about  LL'  is  OA  X  PL.     It  follows  from  the  definition  of 

*  See  Love's  Theory  of  Elasticity,  Vol.  2;  Nature,  Vol.  88,  p.  531  (1912)  for  an  instructive 
paper  by  Prof.  Hopkinson,  on  "The  Pressure  of  a  Blow";  also  Journal  of  the  Franklin  In- 
slitutc,  Vol.  172,  p.  22  (1911)  for  an  account  of  some  determinations  of  the  time  of  impact 
of  metal  spheres. 


238 


Chap,  xi 


the  term,  that  the  angular  momentum  of  a  particle  about  a  line  parallel  to  its 
momentum  is  zero;  and  about  a  line  perpendicular  to  its  momentum  the 
angular  momentum  is  the  product  of  the  momentum  and  the  distance  from 
the  line  to  the  particle. 

There  is  another  method  for  computing  the  angular  momentum  of  a  mov- 
ing particle  about  a  line  which  is  more  simple  generally  than  that  described 
in  the  definition  of  angular  momentum.  It  is  as  follows:  we  resolve  the 
momentum  into  three  rectangular  components,  one  of  which  is  parallel  to  the 
axis  ot  moments  —  then  the  other  two  are  perpendicular  to  the  axis  — ,  and 
add  the  moments  of  the  two  perpendicular  components  about  the  line;  the 
sum  equals  the  angular  momentum  of  the  particle.  Proof:  Imagine  the 
momentum  OC  (Fig.  370)  resolved  first  into  two  rectangular  components  OA 
and  OB  as  before,  and  then  OA  into  any  two  rectangular  components  per- 
pendicular to  LV.  These  last  two  are  not  shown  in  the  figure  but  their 
relations  to  OA  and  the  axis  LL'  are  shown  in  projection  on  the  plane  QQ 
in  Fig.  371.      The  moment  of  the  component  O'M  about  LL'  is  O'AI  X  L'm 

'     B 


U^ 


X 


Fig.  371 


Fig.  372 


=  O'M  X  O'L'  sin  fi  =  O'M  sin  fx  X  O'L'.  The  moment  of  the  component 
O'N  is  O'N  X  L'n  =  O'N  X  O'L'  sin  7  =  O'N  sin  7  X  O'L'.  Hence  the  sum 
of  the  moments  =  (O'M  sin  fi  +  O'N  sin  7)  O'L'  =  O'A'  sin  a  X  O'L'  =  O'A' 
X  O'L'  sin  a  =  O'A'  X  L'P'  which  is  the  angular  momentum  of  the  particle 
as  defined.  *■ 

By  angular  momentum  of  any  collection  of  particles  (body)  about  a  line  is 
meant  the  algebraic  sum  of  the  angular  momentums  of  the  particles  about 
that  line.  In  the  case  of  a  rigid  body  rotating  about  a  fixed  axis,  the  angular 
momentum  of  the  body  about  the  axis  of  rotation  can  be  computed  quite 
easily.  Thus  let  Wi,  fth,  etc.,  =  the  masses  of  the  particles  of  the  body;  r\, 
1-2,  etc.,  =  the  distances  of  the  particles  respectively  from  the  axis  of  rotation; 
and  CO  =  the  angular  velocity  of  the  body.  Then  the  linear  velocities  of  the 
particles  are  respectively  rico,  r2co,  etc.  (Art.  37),  and  their  linear  momentums 
are  Wiriw,  m^roco,  etc.  These  momentums  are  perpendicular  to  the  axis  of 
moments;  hence  the  angular  momentums  are  ntiViuri,  nhrooir-i,  etc.  And  since 
these  are  of  the  same  sign,  the  angular  momentum  of  the  body  is  Wi^i^co  + 
miV^cji  +  .  .  .  =  coSwr-  =  co/,  where  /  =  the  moment  of  inertia  of  the 
body  about  the  axis  of  rotation  (Art.  36). 

A  general  formula  for  the  angular  momentum  of  a  body  about  a  line  can  be 


Art.  48 


239 


arrived  at  as  follows:  Let  P  (Fig.  372)  be  one  of  the  particles  of  the  body, 
OX  the  line  about  which  to  compute  the  angular  momentum,  and  PD  =  the 
velocity  of  P.  Let  OXYZ  be  a  set  of  fixed  coordinate  axes;  x,  j,  and  z  =  the 
(varying)  coordinates  of  P;  m  =  mass  of  P;  d  =  velocity  of  P;  v^c,  Vy,  and  v^  = 
the  axial  components  of  v  (represented  by  PA ,  PB,  and  PC  respectively).  Then 
to  some  scale,  PD  represents  the  momentum  mv  of  the  particle,  and  PA,  PB, 
and  PC  represent  the  axial  components  of  the  momentum;  these  equal  mv^, 
mvy,  and  mv^  respectively.     Hence  the  angular  momentum  of  P  about  OZ 


'J/) 


Y. 


A'^y 


-^-x- 


R> 


is  mVyX  —  mV:,y,  and  the  angular  momentum  of  the  entire  body  is 

2  {mVyX  —  mVxy). 

We  will  now  ascertain  how  the  angular  momentum  of  a  body  about  any 
line  depends  on  the  forces  concerned  in  the  motion.  Let  P,  Fig.  373,  be  one 
of  the  particles  of  a  body,  OX  a  fixed  line  about  which 
the  angular  momentum  in  taken,  R  =  the  resultant  of 
all  the  forces  acting  on  this  particle,  v  =  its  velocity, 
and  a  =  its  acceleration.  Further,  let  the  coordinates  of 
P  at  any  particular  instant  under  consideration  be  .v,  y, 
and  s  referred  to  axes  one  of  which  is  the  line  OX;  R^, 
Ry,  and  R^  =  the  axial  components  of  R;  v^,  Vy,  and 
Vz  =  the  axial  components  of  v;  a^,  Oy,  and  Oz  =  the 
axial  components  of  a;  and  Tz  =  the  torque  of  all  the 
forces  acting  on  P  about  the  z  axis.  Then  T^  =  RyX  -  R^y  (Art.  8);  and 
since  Rx  =  ma^  and  Ry  =  may  (Art.  34), 

Tg  =  niGyX  —  maxy. 

Now  imagine  one  equation  like  the  last  written  down  for  each  particle  of  the 
body.  The  sum  of  the  left-hand  members  equals  the  sum  of  the  right-hand 
members  of  course.  To  the  first  sum  the  internal  forces  (exerted  by  the 
particles  upon  each  other)  contribute  nothing  because  these  internal  forces 
occur  in  pairs,  the  forces  of  each  being  colinear,  ecj^ual,  and  opposite,  and 
so  the  moments  of  such  two  forces  cancel.  Therefore,  the  first  sum  is  also 
the  torque  of  the  external  forces  about  the  2  axis.     Thus,  we  have 


Fig.  373 


SPj  =  2  {mayX  —  maxy), 


(i) 


where  SP^  =  the  torque  of  all  the  external  forces,  acting  on  the  body,  about 
the  2  axis.  The  second  sum,  2  {mayX  —  maxy),  equals  the  rate  at  which  the 
angular  momentum  of  the  body  about  the  z  axis  is  changing.  We  prove  this 
by  differentiating  the  expression  for  angular  momentum  about  the  2  axis, 
S  {mVyX  —  mVxy),  with  respect  to  the  time;   thus 


-J-  2  {mVyX  —  mVxy)  =  2 
at 


m 


(dvy      ,       dx\  (dvx      ,       dyW 


240  Chap,  xi 

Now  dvy/dt  =  ay,  dx/dt  =  Vx,  dvx/dt  =  d,  and  dy/dt  =  Vy-,  and  substitution 
of  these  equivalents  of  the  four  derivatives  in  the  long  equation  gives 

-7- 2  {mVyX  —  mvxy)=  2  {mayX  —  maxj), 
at 

which  was  to  be  proved.  Thus  finally  we  have  the  important  principle  that 
the  torque  oj  the  external  forces,  acting  on  any  body,  about  any  line  equals  the  rate 
at  which  the  angular  momentum  of  the  body  about  that  line  is  changing,  or 

•     2r,  =  dh/dt,  (2) 

where  the  line  in  question  is  called  z,  and  hz  =  the  angular  momentum  of  the 
body  about  that  line. 

For  an  example  we  will  apply  the  foregoing  principle  to  determine  the 
torque  of  the  water  flowing  through  the  water  motor  (Barker  Mill)  repre- 
sented in  Fig.  374.  Essentially,  the  motor  consists  of  a  horizontal  cylinder 
AB,  mounted  on  a  vertical  pivot  C,  and  an  inlet  D  connected  by  a  water-tight 
sleeve  joint  to  a  feed  pipe  E.  On  opposite  sides  of  the  cylinder  and  near  its 
ends  there  are  orifices  or  nozzles  through  which  the  water  escapes  horizontally. 
The  water  turns  the  motor  in  the  opposite  direction.  Let  W  =  the  weight  of 
water  escaping  per  unit  time,  v  =  the  velocity  of  escape  relative  to  the  orifices, 
and  CO  =  the  angular  velocity  of  the  motor.  The  amount  of  water  which 
escapes  in  a  short  interval  of  time  A/  is  W M;  and,  since  the  absolute  velocity 
of  escape  =  t>  —  rco  (Art.  53),  the  angular  momentum  of  this  water  about  the 
axis  of  rotation  is  {WM/g)  (v  —  rw)  r.  Hence  the  rate  at  which  the  motor 
gives  angular  momentum  to  the  water  is 

(W/g)  {v  -  rco)r, 

and  this  equals  the  torque  of  the  motor  on  the  water;  also  the  torque  of  the 
water  on  the  motor.  ' 

If  the  torque  —  about  any  line  —  of  the  external  forces  acting  on  a  body 
equals  zero,  then  the  rate  of  change  of  the  angular  momentum  of  the  body 
about  that  line  equals  zero;  hence,  if  the  torque  remains  zero  for  any  interval 
of  time,  then  the  angular  momentum  remains  constant.  This  is  known  as  the 
principle  of  the  conservation  of  angular  momentum.  It  can  be  well  illustrated 
by  means  of  the  apparatus  on  which  the  man  (Fig.  375)  is  standing.  It  consists 
of  a  metal  plate  A  supported  on  balls  in  suitable  circular  races  in  A  and  B  so 
that  A  can  be  rotated  about  the  line  C  with  very  little  friction  resistance;  B  is 
fixed.  Imagine  that  a  man  has  mounted  the  plate  A  and  holds  a  balancing 
pole  as  shown,  all  being  at  rest;  then  the  angular  momentum  of  the  man-plate- 
pole  system  about  CC  equals  zero.  Now  suppose  that  the  man  exerts  himself 
in  any  way,  to  move  the  pole  about  for  example,  but  touches  nothing  except 
A  and  the  pole.  The  only  external  forces  acting  on  the  system  are  gravity, 
reactions  of  the  balls  on  A ,  and  the  air  pressure.  The  first  has  no  torque  about 
C;  the  other  two  very  little  and  are  negligible  here.     Hence  there  is  no  external 


Art.  48 


241 


torque  about  C,  and  the  angular  momentum  of  the  system  about  C  equals  zero 
always.  This  is  strikingly  illustrated  if  the  man,  without  moving  his  feet  on 
the  plate,  trys  to  rotate  the  pole  (over  his  head  as  shown)  about  C.  In  doing 
so,  he  and  A  begin  to  rotate  in  the  opposite  direction.  If  /  and  /'  =  the 
moments  of  inertia  of  man  (and  .1)  and  the  pole  respectively  about  C,  and  co 
and  w'  =  their  angular  velocities  at  any  instant,  then  the  principle  requires 
that  the  angular  momentums  /w  and  Ico'  shall  be  equal  (and  opposite).  Or, 
imagine  the  man-plate-pole  system  is  given  an  angular  velocity  by  external 
means  (the  man  holding  the  rod  as  shown,  say),  and  then  left  to  itself.  If  now 
the  man  should  change  the  pole  into  a  vertical  position  before  him,  he  would 
reduce  the  moment  of  inertia  of  the  system  (about  C)  very  materially;  and 
since  the  angular  momentum  must  remain  constant,  the  angular  velocity  of  the 
system  would  increase  accordingly. 

The  grand  illustration  of  the  principle  of  conservation  of  angular  momentum 
is  furnished  by  the  solar  system.  The  system  moves  under  the  influence  of 
no  external  forces;  hence  the  angular  momentum  of  the  system  about  any  line 
remains  constant.  The  angular  momentum  about  a  certain  line  through  the 
mass-center  of  the  system  is  greater  than  that  about  any  other  such  line. 
The  Une  is  known  as  the  invariable  axis  of  the  system  —  a  plane  perpendicular 
to  it  as  the  invariable  plane  —  and  ''is  the  nearest  approach  to  an  absolutely 
fixed  direction  yet  known." 

Center  of  Percussion.  —  Fig.  376  represents  a  body  OC  suspended  like  a 
pendulum;  C  is  the  center  of  suspension,  and  C  is  the  center  of  gravity  or  mass- 
center  of  the  body.     Let  R  =  the  reaction  of  the  axle  supporting  the  pendulum. 


J-4 


=  E 


!lD 


I 


k  -  r   >i<-  r  — H 
Fig.  374 


B 


rO 


Ry 


C 


Fig.  376 


and  P  =  the  time  average  force  of  a  blow  applied  as  shown.  In  general,  R 
would  not  be  vertical  during  the  blow;  so  let  R^  and  Ry  =  the  horizontal  and 
vertical  components  of  the  time-average  of  R  during  the  blow.  The  value 
of  R^  depends  not  only  on  the  force  of  the  blow  P  but  also  on  the  arm  of  the 
blow  with  respect  to  the  axis  of  suspension.     It  will  be  shown  presently  that 


242  Chap,  xi 

if  the  arm  has  a  certain  value,  then  Rx  equals  zero.  The  point  Q  in  OC  (ex- 
tended) and  in  the  line  of  action  of  a  blow  applied  as  just  explained  so  that 
there  is  no  component  axle  reaction  parallel  to  the  blow,  is  called  the  center 
of  percussion  of  the  body  for  the  particular  axis  of  suspension.  {Q  is  the 
point  that  was  called  center  of  oscillation  in  Art.  39.)  The  distance  of  the 
center  of  percussion  from  the  axis  of  suspension  equals 

q  =  k'^/c  =  c  -\-k  /c, 

where  k  =  the  radius  of  gyration  of  the  pendulum  about  the  axis  of  suspension, 

c  =  the  distance  from  the  center  of  gravity  to  that  axis,  and  k  =  the  radius 

of  gyration  about  a  Une  through  the  mass-center  and  parallel  to  the  axis  of 

suspension. 

To  develop  the  expression  for  q  given  above  let  M  =  the  mass  of  the  body, 

p  =  the  arm  of  P  about  the  axis  of  suspension,  co  =  the  angular  velocity  of  the 

body  produced  by  the  blow,  and  At  =  the  duration  of  the  blow.     By  the  end 

of  the  blow  the  velocity  of  C  will  be  ceo,  and  practically  horizontal;    hence, 

according  to  Art.  46, 

P  -  Rx  =  Mco:/At. 

The  only  force  which  has  a  torque  about  0  during  the  blow  is  P;  hence 

Pp  =  M¥ui/M. 

These  two  equations  solved  simultaneously  for  R^  give  R^  =  P  (1  —  cp/P); 
therefore,  if  ^  =  k^/c,  Rx  =  o  which  was  to  be  shown. 

Every  American  boy  has  batted  a  baseball  a  few  times  in  such  a  way  that 
the  bat  "stung  "  his  hands;  and  he  soon  learned  that  such  stinging  is  a  result 
of  impact  near  his  hands  or  quite  near  the  big  end  of  the  bat;  in  fact,  quite 
remote  from  the  center  of  percussion  of  the  bat  (with  reference  to  the  particular 
axis  of  rotation  about  which  the  bat  was  being  swung  at  the  instant  of  impact). 
Such  a  blow  also  results  in  rapid  vibrations  of  the  material  of  the  bat  which 
cause  the  sting.  Large  pendulums  are  used  in  certain  impact  testing  machines 
for  striking  a  blow.  To  avoid  the  impulsive  reaction  at  the  suspension  and 
vibrations  in  the  pendulum,  they  are  always  so  arranged  that  the  line  of  action 
of  the  blow  passes  through  the  center  of  percussion  of  the  pendulum. 

§  2.  Angular  Impulse.  —  If  the  line  of  action  of  a  force  is  fixed  in  posi- 
tion then  the  angular  impulse  of  that  force  for  any  interval  about  any  line  is 
the  moment  of  the  impulse  of  the  force  for  the  interval  about  that  Hne.  The 
moment  of  an  impulse  is  computed  just  like  moment  of  a  force  (Art.  8)  or 
angular  momentum;  that  is,  we  resolve  the  impulse  into  two  components, 
one  parallel  and  one  perpendicular  to  the  line  and  then  we  take  the  product 
of  the  perpendicular  component  and  the  distance  from  it  to  the  line.  If  the 
line  of  action  of  the  force  changes  then  the  angular  impulse  of  the  force  about 
any  line  for  any  interval  is  the  algebraic  sum  of  the  angular  impulses  for  all 
the  elementary  portions  of  time  which  comprise  the  interval.  Thus  let  F  = 
the  force,  i"  —  /'  =  the  interval,  6  =  the  angle  between  the  line  of  action  F 


Art.  48  243 

and  the  line,  and  p  =  the  perpendicular  distance  between  the  two  lines.     Then 

the  angular  impulse  is 

Xt"  nt" 

F  dt-?,md-  p=   \      F  sin  d  -  p  dt. 

Since  F  sinO  •  p  —  the  torque  of  the  force  about  the  line  in  question,  the 
angular  impulse  of  the  force  may  also  be  regarded  as  the  time-integral  of  the 
torque  of  the  force.  Hence,  if  T  —  the  torque  of  the  force  about  the  line  at 
any  instant  then  the  angular  momentum  for  the  interval  equals 

Tdt. 


£ 


h' 
Now  let  us  integrate  equation  (2)  over  any  interval  /"  —  /'  say;   then 

r  ^T,dt,  or 2    r   T,dt,  =  ///'  -  hj  =  A/?„  (3) 

in  which  JiJ  and  lij'  denote  the  angular  momentums  of  the  body  about  the 
z  axis  at  the  times  t'  and  t"  respectively.  Equation  (3)  can  be  put  into  the  fol- 
lowing principle  of  angular  impulse  and  momentum:  The  sum  of  the  angular 
impulses  of  all  the  external  forces  acting  on  a  body  about  any  line  equals  the 
increment  in  the  angular  momentum  of  the  body  about  that  line. 

49.  Gyrostat 

§  I.  General  Description.  —  The  words  gyroscope  and  gyrostat  are 
generally  used  synonymously  but  sometimes  a  distinction  is  made,  as  follows: 
A  gyrostat  consists  of  a  wheel  and  axle,  both  being  symmetrical  to  the  axis  of 
the  axle,  and  mounted  so  that  they  may  be  rotated  about  that  axis;  a  gyro- 
scope consists  of  a  gyrostat  mounted  in  a  frame  which  can  be  rotated.  Fig. 
377  represents  a  common  form  of  gyroscope;  the  gyrostat  (wheel  W  and  axle 
A  A')  is  supported  by  a  ring  R  which  can  be  rotated 
about  the  axis  BB';  the  axle  BB'  is  supported  by  the  /^^^-■^ 

forked  pillar  F  which  can  be  rotated  about  the  axis  CC.    l\^5^/\^^.    . 
Thus  the  wheel   can  be  rotated  about  its  center  into      t^^^^&A^^ 
any  desired  position.     The  gyroscope   seems  to  have        ^   IWV>^-/^)"^b' 
been  designed  for  illustrating  principles  of  composition  ^^^y^JJJ 

of  rotations  (Art.  54).     In  1852  Foucault  (French  phy-  x.^-^^ 

sicist)  made  an  interesting  application   of  the  instru-  —  i-"^-s^i=P" 

ment;    by   its    means   he  practically  made  visible  the  k:;--_1_--^ 

rotation  of  the  earth.     More  recently  the  gyroscope  has  p^^ 

been  made  use  of  in  several  connections,  —  to  steer  a 
torpedo,  to  serve  as  a  substitute,  unaffected  by  the  iron  of  the  ship,  for  the 
ordinary  (magnetic)  mariner's  compass,  to  stabilize  a  mono-rail  car,  and  to 
steady  a  ship  in  a  rough  sea;   it  has  been  proposed  also  to  stabiHze  flying 
machines  by  means  of  a  gyroscope. 

When  its  wheel  is  spinning,  a  gyroscope  possesses  properties  which  seem 


244 


Chap,  xi 


peculiar  to  students  as  yet  uninformed  in  the  matter,  inasmuch  as  it  does  not 
always  respond  as  expected  to  efforts  made  to  change  its  motion  or  position. 
For  example,  if  a  gyroscope  like  that  represented  in  Fig.  377,  well  made  and 
practically  frictionless  at  all  bearings  and  pivots,  be  grasped  by  the  pillar  and 
then  moved  about  in  any  way,  the  axle  of  the  wheel  remains  fixed  in  direction 
in  spite  of  any  attempt  to  alter  it.  The  (gimbal)  method  of  support  makes 
it  impossible  to  exert  any  resultant  torque  on  the  gyrostat  (by  way  of  the 
pillar)  about  any  line  through  the  center;  and  hence,  as  will  be  proved  later, 
the  direction  of  the  axle  cannot  be  thus  changed.  It  is  this  property  of  per- 
manence of  direction  of  the  spin-axis  of  a  gimbal-supported  gyrostat  which  is 
made  use  of  in  the  self-steering  torpedo. 

For  another  example,  consider  the  effect  of  a  torque  applied  directly  to  the 
gyrostat.  A  vertical  force,  say,  applied  at  A  would  turn  the  gyrostat  when 
not  spinning  about  the  axis  B.  But  when  spinning,  that  force  U  would  rotate 
the  spin-axis  about  the  axis  C,  the  direction  of  rotation  depending  upon  the 
direction  of  spin.  When  the  gyrostat  is  spinning  in  the  direction  indicated 
by  the  arrow  co,  then  such  force  U  would  rotate  the  spin-axis  about  C  in  the 
direction  indicated  by  the  arrow  fx.  Again,  a  horizontal  force  applied  at  A, 
say,  would  turn  the  gyrostat  when  not  spinning  about  the  axis  C.  But  w^hen 
spinning,  such  force  L  would  rotate  the  spin-axis  about  BB';  and  in  the  direc- 
tion indicated  by  the  arrow  X  if  the  spin  is  as  indicated.  This  behavior  of  a 
spinning  gyrostat  under  the  action  of  torque  is  exhibited  more  strikingly  by 
a  gyroscope  represented  plainly  in  Fig.  378.     The  wheel  may  be  spun  on  the 

axle  A;  the  gyrostat  and  its 
frame  may  be  rotated  about 
the  axis  BB';  and  all  may  be 
rotated  about  the  axis  CC. 
W  is  the  weight  which  can  be 
clamped  on  the  stem  A'  to 
balance  or  unbalance  the  frame 
with  respect  to  the  axis  BB'. 
Now  imagine  W  clamped  so 
that  the  frame  (with  W  and 
the  gyrostat)  is  unbalanced. 
Then  if  the  gyrostat  is  set  spinning  and  the  frame  be  released  in  the  position 
shown,  say,  the  frame  will  not  rotate  about  BB'  but  about  CC.  The 
direction  of  this  rotation  depends  on  the  direction  of  spin  and  on  the  di- 
rection of  the  torque  of  gravity  about  BB'.  If,  for  example,  W  is  clamped 
quite  near  BB'  so  that  the  torque  of  gravity  is  clockwise  as  seen  from  B  and 
the  spin  is  as  indicated,  then  A  rotates  toward  B.  This  rotation  persists  except 
in  so  far  as  it  is  interfered  with  by  friction  at  the  pivots,  and  air  resistance. 
We  might  recite  still  other  peculiar  performances  of  a  gyrostat  but  the  fore- 
going suffice  for  our  purpose.  Professor  Perry's  book  on  "Spinning  Tops" 
would  be  found  interesting  in  this  connection. 


Fig.  378 


Art.  49  245 

Any  such  rotation  of  the  axis  of  a  spinning  gyrostat  is  called  a  precessional 
motion  or  precession  of  the  axis  or  of  the  gyrostat;  the  axis  and  the  gyrostat 
are  said  to  precess.  We  will  call  precession  normal  or  obiique  according  as 
the  axis  precesses  about  a  line  perpendicular  or  inclined  to  the  axis.  It  may 
not  be  clear  from  the  foregoing  examples  of  precession  how  to  predict  the 
direction  of  precession  that  would  result  by  applying  a  given  torque  to  a  gyro- 
stat with  a  given  spin.  The  following  is  a  simple  rule  for  predicting;  it  is 
based  on  the  dynamics  of  the  whole  matter  as  will  be  seen  later:  "  When  forces 
act  upon  a  spinning  body  tending  to  cause  rotation  about  any  other  axis  than 
the  spinning  axis,  the  spinning  axis  sets  itself  in  better  agreement  with  the  new 
(other)  axis  of  rotation;  perfect  agreement  would  mean  perfect  parallelism, 
the  direction  of  rotation  being  the  same."  (From  "Spinning  Tops".)  Or, 
what  amounts  to  the  same  thing,  the  precession  is  such  as  to  turn  the  spin- 
vector*  toward  the  couple  or  torque-vector. 

The  following  is  an  incomplete  proof  of  the  foregoing  rule.  Further  ex- 
planation is  given  in  the  next  section  and  in  Art.  56.  Fig.  379  represents 
a  gyrostat  pivoted  at  O  so  that  it  can  be 
rotated  freely  about  that  point;  we  sup- 
pose the  center  of  gravity  of  the  gyrostat 
to  be  at  O.     Imagine  that  the  gyrostat  is    -TIX 

at  rest,  not  spinning,  in  the  position  shown,  y^i^ ~^^^ 

and  that  a  downward  force  is  applied  to        ^ 
the   axle  on  the  left-hand  side  of  0  and 

downward.  The  torque  makes  the  gyrostat  rotate  about  the  axis  OB,  that  is 
the  torque  produces  angular  momentum  about  that  axis.  The  amount  of  an- 
gular momentum  produced  is  proportional  to  the  torque  and  to  the  duration  of 
its  action  (see  Art.  48).  This  angular  momentum  may  be  represented  by  a 
vector  on  OB,  the  length  of  the  vector  representing  the  amount  of  the  angular 
momentum  and  the  arrow-head  pointing  so  as  to  agree  with  the  direction  of 
rotation,  according  to  the  usual  convention,  that  is,  forward  in  this  case.  Now 
imagine  that  the  axis  of  the  gyrostat  is  at  rest  in  the  position  shown  but  the 
wheel  spinning,  say,  counter-clockwise  when  viewed  from  the  right.  The  an- 
gular momentum  of  the  spinning  gyrostat  about  its  axisf  would  be  represented 
by  a  vector  on  OA  pointing  in  the  direction  OA;    let  01  be  that  vector. 

*  A  spin-vector  is  a  vector  on  the  axis  of  spin,  its  arrow-head  pointing  to  the  place  from 
which  the  spin  appears  counter-clockwise;  or  —  what  amounts  to  the  same  thing  —  the  arrow- 
head points  in  the  direction  along  which  the  axis  would  advance  if  it  were  a  right-hand  screw 
turning  in  a  fixed  nut.  The  length  of  the  vector  —  immaterial  in  the  present  connection  — 
represents  the  angular  velocity  of  spin  to  some  convenient  scale.  Likewise  the  couple-vector 
(see  Art.  8)  is  a  vector  perpendicular  to  the  plane  of  the  couple  pointing  to  the  place  from  which 
the  rotation,  which  the  couple  tends  to  produce,  would  appear  counter-clockwise;  or  —  what 
amounts  to  the  same  thing  —  the  arrow-head  points  in  the  direction  along  which  the  vector 
would  advance  if  it  were  a  right-handed  screw  turned  by  the  couple  in  a  fixed  nut. 

t  This  angular  momentum  is  greater  than  that  for  any  other  line,  and  hence  may  bf 
regarded  as  the  total  or  resultant  angular  momentum  of  the  gyrostat  (see  Art.  55). 


246 


Chap,  xi 


Now  suppose  that  the  torque  already  described  comes  into  action,  and  let  OJ 
represent  the  angular  momentum  which  it  would  produce  in  a  short  interval 
of  time.  This  angular  momentum  added  to  the  original  angular  momentum 
gives  OR  as  the  resultant  angular  momentum  of  the  gyrostat  at  the  end  of  the 
interval.  It  seems,  therefore,  that  the  spin-axis  would  coincide  with  OR  at 
the  end  of  the  interval;  indeed,  that  axis  does  approach  OR,  that  is  the  spin- 
axis  turns  toward  the  torque-axis  as  stated  in  the  rule  which  v/e  undertook 
to  prove. 

The  approach  just  mentioned  is  not  a  direct  one;  the  gyrostat  yields  slightly 
to  the  torque  just  as  though  there  were  no  spin;  that  is  the  wheel  rises  (in  this 
instance)  slightly.  This  is  only  the  first  (small)  swing  of  a  rapid  oscillation 
of  the  spin-axis  —  nutation  as  it  is  called  —  which  accompanies  the  (more 
prominent)  precession  of  the  spin-axis  toward  the  torque-axis.  The  (unavoid- 
able) friction  at  the  pivot  O  rapidly  damps  this  oscillation  so  that  the  oscilla- 
tion generally  escapes  notice.  The  mentioned  rise  of  the  spin-axis  may  be 
explained  as  follows:  In  the  approach  of  that  axis  toward  OR  the  gyrostat 
rotates  about  OC,  due  to  which  it  acquires  angular  momentum  about  OC, 
clockwise  when  viewed  from  above;  but  since  there  is  no  torque  about  OC, 
the  gyrostat  can  acquire  no  (resultant)  angular  momentum  about  that  line 
(see  Art.  48  on  conservation) ;  hence  the  spin-axis  rises  so  that  at  each  instant 
the  component  along  OC  of  the  angular  momentum  due  to  spin  just  equals  the 
angular  momentum  due  to  the  rotation  about  OC. 

There  is  another  item  of  gyrostat  behavior  worth  noting  here.  Suppose 
that  the  gyrostat  shown  in  Fig.  378  to  be  precessing  as  already  explained. 
If  the  precession  be  hurried,  say  by  means  of  a  horizontal  push  applied  at  .4', 
the  center  of  gravity  of  the  frame  (with  gyrostat  and  weight)  rises;  if  the 
precession  be  retarded,  the  center  of  gravity  descends.  This  behavior  is  in 
accordance  with  the  rule  for  predicting  precession.  In  the  first  case  we  have 
a  torque  about  CC;  the  torque  vector  is  in  the  direction  OC;  the  spin- vector 
is  in  the  direction  OA';  and  in  accordance  with  the  rule  OA'  turns  toward  OC , 
that  is  the  center  of  gravity  rises.  In  the  second  case  we  have  a  torque  about 
CC  but  the  torque- vector  is  OC;  and  the  spin- vector  OA'  turns  toward  that 
vector,  that  is  the  center  of  gravity  descends.  Thus  we  may  state  as  another 
rule:  Hurry  a  precession,  the  gyrostat  rises  or  opposes  the  torque  which  causes 
the  precession;  retard  a  precession,  the  gyrostat  falls,  or  yields  to  the  torque 
which  causes  the  precession. 

Self -steering  Torpedo.  —  The  gyroscope  of  such  a  torpedo  is  linked  to  appro- 
priate valves  of  a  compressed  air  engine  in  such  a  way  that  any  turning  of 
the  spin- axis  toward  either  side  of  the  torpedo  causes  the  engine  to  turn  the 
(vertical)  rudder  of  the  torpedo  in  the  opposite  direction.  Prior  to  projection 
of  a  torpedo,  the  gimbals  are  locked  so  as  to  hold  the  spin-axis  of  the  gyrostat 
parallel  (or  inchned  at  any  desired  angle)  to  the  axis  of  the  torpedo.  During 
the  discharge  of  the  torpedo,  the  gyrostat  is  automatically  set  spinning  and 
the  gimbals  are  unlocked.     During  the  flight,  the  spin-axis  continues  to  point 


Art.  49  247 

in  its  original  direction ;  any  deviation  of  the  torpedo  from  its  intended  course 
changes  the  incHnation  of  the  spin-axis  relative  to  the  torpedo;  simultaneously 
the  gyroscope  actuates  the  rudder  as  explained,  and  the  torpedo  is  deflected 
back  toward  its  proper  direction.  Like  a  common  pendulum  swinging  to  its 
lowest  position,  the  torpedo  swings  beyond  a  mean  direction,  and  is  then  swung 
back  again  by  the  rudder.  And  this  oscillation  is  kept  up  during  the  flight  so 
that  the  actual  path  of  the  torpedo  is  a  zigzag,  about  two  feet  wide.  A  gyro- 
stat (wheel  and  axle)  weighing  2  pounds  and  rotating  at  2500  revolutions  per 
minute  has  been  made  to  serve  the  purpose  just  described. 

Gyro-compass.  —  For  our  purpose  we  may  regard  a  gyro-compass  as  con- 
sisting essentially  of  a  gyrostat  (wheel  and  axle),  the  axle  supported  in  a  ring 
or  case,  and  the  ring  suspended  from  above.  See  A,  Fig.  380.  Such  a  com- 
pass, when  the  gyrostat  is  spinning,  sets  its  spin-axis  into  the  plane  of  the 
meridian  at  the  place  where  the  compass  happens 
to  be.  Imagine  such  a  compass  to  be  set  up  at  the 
equator  with  its  spin-axis  pointing  east  and  west,  and 
suppose  that  the  direction  of  spin  is  counter-clockwise 
when  viewed  from  the  west.  The  rotating  earth 
carries  the  gyrostat  eastward;  the  spin-axis  would 
remain  parallel  to  its  original  position  if  the  gyrostat 
were  supported  in  frictionless  gimbals,  and  would  in 
time  be  positioned  as  shown  at  B.  Now  consider  the  gyrostat  as  shown  at 
B,  supported  not  in  gimbals  but  suspended  from  above  as  in  the  gyro- 
compass. The  supporting  force  (above)  and  the  force  of  gravity  would  have 
a  torque  counter-clockwise  as  viewed  from  the  north;  thus  the  torque  vector 
would  point  toward  the  reader.  The  spin- vector  points  to  the  right;  hence 
the  torque  would  turn  the  end  of  the  spin-axis  marked  n  from  the  west 
toward  the  north. 

Of  course  the  action  is  not  precisely  as  outUned  above,  that  is  the  spin-axis 
does  not  remain  parallel  to  its  original  position  for  a  time  and  then  yield  to 
the  influence  of  the  torque  mentioned.  The  action  is  really  continuous;  the 
slightest  rotation  of  the  compass  with  the  earth  from  the  position  A  induces 
the  gravity  torque,  and  the  spin-axis  begins  to  turn  toward  the  meridian  as 
described. 

Though  the  restraint  of  the  support  (fine  wire  in  the  Sperry  and  mercury 
float  in  the  Anschutz  compass)  is  very  small,  the  gravity  torque  is  so  small 
that  the  turning  of  the  spin-axis  into  the  meridian  is  very  slow.  Like  a 
magnetic  compass  the  gyro-compass  swings  beyond  the  meridian  from  a 
deflected  position  and  oscillates  for  a  time.  In  the  Anschutz  type  the  period 
of  a  free  oscillation  is  about  i  hour  and  20  minutes.  Special  damping  ar- 
rangements reduce  the  oscillations  to  zero  (from  a  deflected  position  of  40 
degrees)  in  about  one  and  one-half  hours.  The  spin  is  maintained  electrically, 
at  about  20,000  revolutions  per  minute. 

Mono-rail  Car.  —  A  car  on  a  single  rail  can  be  rendered  stable  even  if  the 


248 


Chap,  xi 


center  of  gravity  of  the  car  is  above  the  rail  by  means  of  a  suitable  gyroscope 
apparatus.  Fig.  381  represents  the  germ  of  one  type  of  such  apparatus. 
A  A'  is  the  spin-axis,  Z,  is  a  lever  rigidly  fastened  to  the  axle  BB'  by  means  of 
which  the  gyrostat  can  be  made  to  precess  about  BB'.  Imagine  the  car  to  be 
standing  or  travelling  in  an  upright  position,  the  gyrostat  spinning,  and  a  man 
standing  on  the  car  so  that  he  may  grasp  and  operate  the  lever.  Now  suppose 
that  the  car  is  tilted,  as  by  a  wind  against  either  side.  The  car  exerts  tilting 
forces  on  the  gyrostat  axle  at  B  and  B',  the  torque-vector  of  which  is  parallel 


A' 


•  =c= 

4 

r>B' 

cM               t^.c 

Fig.  381 


^s^^^^ 


to  the  rail;  hence  (see  the  stated  rule)  the  spin-axis  begins  to  set  itself  parallel 
to  the  rail,  that  is  it  precesses  about  BB'.  The  axle  BB'  exerts  (righting) 
reactions  on  the  car  but  if  the  man  will  hurry  the  precession,  the  (heavy, 
rapidly  spinning)  gyrostat  will  rise  against  the  tilting  forces  and  carry  the  car 
back  with  it  toward  the  vertical  position.  It  is  conceivable  that  a  skillful 
operator  could  put  the  car  back  into  its  vertical  position  in  one  swing,  but  in 
general  he  would  swing  the  car  beyond  the  vertical,  then  back  again  and  after 
a  few  oscillations,  into  its  vertical  position. 

Gyro-stabilizers  as  now  built  automatically  perform  the  function  of  the  man 
of  the  preceding  explanation,  and  they  include  two  gyrostats,  spinning  in 
opposite  directions,  to  enable  the  car  to  run  on  a  ciu-ve.  The  gyrostat  wheels 
of  a  certain  Brennan  mono-rail  car  (40  feet  long  and  weighing  22  tons)  are  3I 
feet  in  diameter;  each  weighs  f  tons,  and  spins  at  3000  revolutions  per  minute 
(in  a  vacuum  to  avoid  air  friction).  Such  a  car  has  taken  curves  of  105  feet 
radius  at  a  speed  of  7  miles  per  hour  without  appreciable  disturbance  of  the 
level  of  the  car  floor.  The  spin  is  maintained  by  electric  means;  in  fact  each 
gyro-wheel  is  made  the  armature  of  a  motor  and  this  is  driven  by  a  generator 
on  the  car. 

Schlick  Gyro-stabilizer  for  Reducing  the  Rolling  of  a  Ship.  —  This  is  repre- 
sented in  Fig.  382.  The  gyrostat  is  mounted  in  a  rigid  frame  F  which  is  sup- 
ported in  bearings  B  and  B'  fixed  on  the  ship.  Thus  the  wheel  can  be  spun 
about  A  A'  and  the  axle  A  A'  can  precess  about  BB'.  P  is  a  brake  pulley  by 
means  of  which  this  precession  can  be  controlled.  Explanation  of  the  steady- 
ing action  of  this  device  is  beyond  the  scope  of  this  article.  Such  a  stabilizer 
has  been  tried  out  in  a  ship  no  feet  long,  12  feet  wide  and  of  58  tons  displace- 
ment.    The  gyro- wheel  weighed  iioo  pounds,  was  i  meter  in  diameter,  and 


Art.  49  249 

was  spun  at  1600  revolutions  per  minute.  In  still  water  the  ship  would  settle 
down  from  a  heel  of  20  degrees  to  one  of  |  degree  in  about  20  single  oscillations; 
the  period  was  about  4I  seconds.  The  stabilizer  produced  the  same  extinction 
in  less  than  three  oscillations  of  6  seconds  period.  (See  London  Engineering, 
Vol.  83,  p.  448  (1907)). 

§2.  Rate  or  Normal  Precession;  Determination  of  Forces. — In 
the  preceding  section,  we  discussed  the  effect  of  a  torque  on  a  spinning  gyro- 
stat in  a  qualitative  way;  we  will  now  discuss  the  matter  quantitatively. 
Let  /  =  the  moment  of  inertia  of  the  gyrostat  about  the  axis  of  spin  and 
CO  =  the  angular  velocity  of  spin;  then  /co  —  the  angular  momentum  of  the 
gyrostat  about  that  axis  (Art.  48).  If  T  =  the  applied  torque,  the  angular 
momentum  produced  by  it  in  the  element  of  time  dt  is  T  dt,  and  the  angular 
approach  of  the  spin-axis  toward  the  torque-axis  in  that  time  is  lOR  (Fig. 
379)  =  tan-i  (r  dt/Iw)  =  T  dt/Iw.  The  rate  at  which  this  angle  is  described, 
that  is  the  angular  velocity  of  precession  —  generally  denoted  by  O  —  is 

n  =  {IOR)/dt  =  r//co. 

If  the  torque  is  applied  so  that  its  vector  is  always  perpendicular  to  the  axis 
of  spin  OA,  then  there  is  no  torque  about  OA  and  hence  co  is  constant;  if  also 
the  magnitude  of  the  torque  is  constant,  then  it  follows  from  the  preceding 
formula  that  O  is  constant.  That  is,  in  the  case  assumed,  the  velocities  of 
spin  and  precession  are  constant.  The  case  is  quite  analogous  to  that  of  a 
moving  particle  subjected  to  a  constant  force  whose  line  of  action  is  always 
perpendicular  to  the  direction  of  motion  and  in  a  given  plane.  Such  a  force 
does  not  change  the  magnitude  of  the  velocity  but  continually  changes  the 
direction  of  it;  indeed,  the  particle  describes  a  circle  with  constant  speed 
(Art.  34).  Let  P  (Fig.  383)  be  the  particle,  m  =  its  mass,  v  =  its  velocity, 
F  =  the  force,  PQ  be  the  path  and  r  =  the  radius  of  the 

circle.      The   linear   momentum  =  mv;    the   angle.  POQ    -   i.-ja^'^  > ^^I 

through  which  the  vector  mv  is  turned  in  any  time  /  is      ,|__7_t^v$r:r^R 

vt/r.     Since  r  =  mv^/F  (see  Art.  34),  the  angle  =  tF/mv. 

Hence  the  rate  at  which  F  turns  the  linear  momentum 

vector  is  F/mv,  a  result  strictly  analogous  with  T/Ioj,  the 

rate  at  which  the  torque  T  turns  the  angular  momentum  ^^^     ^ 

vector  /o).     The  result  can  be  arrived  at,  independently 

of  Art.  34,  in  a  way  to  bring  out  the  analogy  still  more.     We  may  regard  F 

constant  in  direction  for  an  element  of  time  dt.     During  that  time  it  produces 

an  amount  of  momentum,  in  its  own  direction  PO,  equal  to  F  dt.     Let  PJ 

represent  this  momentum  and  PI  the  initial  momentum  mv.     At  the  end  of 

the  interval  the  (resultant)  momentum  is  represented  by  PR.     Hence  the 

change  in  the  direction  of  the  momentum  is  I  PR  =  {F  dt)  -=-  (mv),  and  the 

rate  at  which  the  change  occurs  is  the  change  divided  by  dt,  that  is  F/mv. 

The  Forces  Acting  on  a  Gyrostat  Precessing  Normally  at  Constant  Speed.  — 
We  will  now  determine  certain  conditions  which  the  forces  in  such  a  case 


250 


Chap,  xi 


always  fulfill.  Incidentally,  we  give  an  alternative  derivation  of  the  formula 
12  =  T/Iw.  We  take  the  gyrostat  represented  by  two  projections  in  Fig.  384. 
^A^  is  the  axis  of  spin,  the  perpendicular  to  the  paper  at  0  is  the  axis  of  pre- 
cession, and  Q  is  the  mass-center  of  the  gyrostat.  The  assumed  directions  of 
spin  and  precession  are  indicated  by  the  curved  arrows  oj  and  12  respectively. 


Fig.  384 


For  the  investigation  we  shall  use  two  sets  of  coordinate  axes,  one  fixed  and 
one  moving.  The  fixed  set  is  OX,  OY ,  and  OZ,  the  latter  not  shown;  OZ  is 
taken  coincident  with  the  precession-axis,  and  OX  and  OY  in  the  plane  in 
.which  the  spin-axis  moves.  The  moving  set  consists  of  NA,  NB,  and  NC; 
NA  is  the  spin-axis  (as  already  stated),  NB  is  the  common  perpendicular  to 
the  axes  of  spin  and  precession,  and  NC  is  perpendicular  to  iV.4  and  NB. 
Let  /'  =  the  moment  of  inertia  of  the  gyrostat  about  the  axis  NC,  e  =  the 
distance  (ON)  between  the  axes  of  spin  and  precession,  </>  =  the  (varying) 
angle  which  the  spin-axis  makes  with  OX,  P  be  any  particle  of  the  gyrostat, 
m  =  its  mass,  r  —  its  distance  from  the  axis  of  spin,  6  =  the  (changing)  angle 
BNP,  a,  b,  and  c  =  the  co()rdinates  of  P  with  respect  to  the  moving  axes,  and 
X,  y,  and  2  =  its  coordinates  with  respect  to  the  fixed  axes. 

It  follows  from  the  trigonometric  relations  in  the  figure  that 

X  =  a  cos  4)  —  (b  -{■  e)  sin  0  =  a  cos  4>  —  r  cos  0  sin  </>  —  e  sin  0, 
y  =  c  sin  0  +  (6  -f  e)  cos  </>  =  a  sin  </>  4- ''  cos  <^  cos  6  -\-  e  cos  <^, 
and       z  =  c  =  f  sin  ^. 

Dififerentiating  these  expressions  with  respect  to  time  (and  noting  that  a,  r, 
and  e  are  constants,  and  that  dd/dt  =  w  and  dip/df  =  12),  we  get  the  following 
values  of  the  x,  y,  and  z  components  of  the  velocity  of  P: 

Vx  =  {cw  —  al2)  sin  ^  —  (6  -f  e)  12  cos  0, 
Vy  =  (al2  —  ceo)  cos  0  —  (6  +  e)  12  sin  0, 
and  Vn  =  b(j3. 

The  angular  momentums  of  P  about  the  axes  OX,  OY,  and  OZ  respectively 
are  (see  Art.  48) 

in{vzy  —  VyZ),     m(vxZ  —  Vzx),     and   m(vyX  —  Vxy). 


Art.  49  ^5^ 

If  we  substitute  in  these  expressions  for  V:,,  Vy,  and  v^,  their  values  as  just  de- 
duced, then  sum  up  for  all  the  particles  of  the  gyrostat,  we  arrive  at  the  follow- 
ing simple  expressions  for  the  angular  momentums  of  the  gyrostat  about  the 
X,  y,  and  z  axes  respectively  :* 

hx  =  loi  cos  0,    hy  =  lu  sin  <^,    and  hz  =  /'O. 

Differentiating  these  expressions  for  h^,,  hy,  and  h,  with  respect  to  time  (and 
remembering  that  co  and  9,  are  assumed  to  be  constant),  we  find  that  the  rates 
at  which  the  angular  momentums  change  are 

dhx/dt  =  —  /a;12  sin  (/>,     dhy/dt  =  Io£l  cos  0,     and    dhjdt  =  o. 

Now  consider  the  instant,  or  position  of  the  gyrostat,  when  the  spin-axis  NA 
is  parallel  to  the  x  axis.  Then  <t>  =  o,  and  the  rates  respectively  equal  o, 
/wS],  and  o;  these  respectively  are  also  the  torques  {T^,  Ty,  and  T^  of  the  ex- 
ternal forces  about  the  %,  y,  and  z  axes,  when  </>  =  o  (Art.  48) ;  that  is 

Tx  =  o,     Ty  =  Ioi%    and     Tz  =  o. 

By  means  of  these  results  and  the  following  paragraph,  it  is  shown  in  §  3 

that 

T,  =  0,     Tp  =  o,    and     T  =  /cofi,  (i) 

where  Ts,  Tp,  and  T  denote  the  torques  of  all  the  external  forces  about  the 
axis  of  spin,  the  axis  of  precession,  and  their  common  perpendicular. 

For  further  information,  we  will  now  resort  to  the  principle  of  the  motion 
of  the  mass-center  (Art.  34,  page  159).  The  mass-center  describes  a  circle 
at  constant  speed;  hence  the  acceleration  of  that  point  is  always  directed 
from  the  mass- center  to  the  center  of  the  circle,  and  its  value  is  rli^  where  r 
denotes  radius  of  the  circle.  Now  let  M  =  mass  of  the  gyrostat,  Rr,  Rp,  and 
R3  =  the  sums  of  the  components  of  all  the  external  forces  along  the  radius, 
the  precession  axis,  and  the  perpendicular  to  these  two  lines;  then  according 
to  the  principle  named  above 

Rr  =  MrO^,     Rp  =  o,     and    i?3  =  o.  (2) 

The  six  equations  in  (i)  and  (2)  are  the  certain  conditions  referred  to  at  the 
bottom  of  page  249;  they  are  applied  in  the  following 

Examples.  —  (i)  Fig.  385  represents  a  side  and  end  view  of  the  armature 
of  the  motor  of  an  electric  locomotive.  The  armature  shaft  is  parallel  to  the 
ties  of  the  track.  We  will  discuss  the  forces  acting  on  the  armature  when  the 
locomotive  is  rounding  a  curve.  Inasmuch  as  we  are  not  now  concerned  with 
the  driving  of  the  locomotive  by  this  motor  we  will  assume  that  the  armature 
is  spinning  but  under  no  load,  the  locomotive  being  driven  around  the  curve 
by  another  locomotive.  And  for  simplicity,  we  assume  that  there  is  no  eleva- 
tion of  the  outer  rail,  so  that  the  precession  of  the  armature  is  normal;  that  is, 

*  In  reducing  the  summations,  the  student  should  note  that 

•Sm(b'-  +  c'-)  =  I,     ^mia'  +  h"-)  =  I',  and 
Sm&c  =  I,mca  =  T^mab  =  ^mb  =  '^mc  =  o. 


252 


Chap,  xi 


we  take  the  angle  between  the  axis  of  spin  and  the  (vertical)  precession-axis 
to  be  90  degrees.  We  take  the  weight  of  the  armature  =  8000  pounds,  its 
radius   of   gyration  =15   inches,   its   speed  =750   revolutions  per   minute, 


1 

1 

-- 

h  — 

/ 

^^000/b^-^ 

Q,=0/P,  =  Q,= 


Fig.  385 


W 


P 

Fig.  386 


distance  between  centers  of  bearings  =  4  feet,  the  radius  of  the  curve  =  2000 

feet,  and  the  speed  of  the  car  =  30  miles  per  hour  (=  44  feet  per  second). 

Then 

/  =  (8000/32.2)  (15/12)2  =  388  slug-feet,2 
w  =  750  X  2  7r/6o  =  78.54  radians  per  second, 

and  fi  =  44/2000  =  0.022  radians  per  second. 

The  forces  acting  on  the  armature  are  gravity  and  the  reactions  P  and  Q  of  the 
bearings  on  the  armature  shaft.  We  neglect  axle  friction  and  imagine  each 
reaction  resolved  into  three  components,  vertical,  parallel  to  the  armature 
shaft,  and  parallel  to  the  rails.  We  distinguish  these  components  by  the  sub- 
scripts 1,2,  and  3,  respectively  (see  the  figure).  If  the  center  of  the  curve 
is  on  the  right,  then  evidently  the  armature  presses  outward  against  the 
bearing  P  and  hence  Qo  =  o.  Since  the  sum  of  the  component  forces 
along  the  rails  =  o,  P3  and  Q3  must  be  equal  and  opposite,  or  else  equal  zero. 
Since  the  torque  about  the  axis  of  precession  must  =  o,  P3  and  Q3  =  o. 
According  to  equation  (2), 

P2  =  (8000/32.2)  (44-/2000)  =  240  pounds. 

The  torque  of  all  the  forces  acting  on  the  gyrostat  about  the  common  per- 
pendicular to  the  spin  and  precession  axes  equals 

IwQ  =  388  X  78.54  X  0.022  =  670  foot-pounds. 

If  the  direction  of  spin  is  the  same  as  the  direction  of  rotation  of  the  car  wheels, 
then  the  torque  is  clockwise  seen  from  the  rear;   hence 

Pi(2ooo  +  2)  +  Qi(2ooo  —  2)  —  8000  X  2000  =  670. 

We  have  also  Pi  -\-  Qi  =  8000;  hence,  solving  these  two  equations  simul- 
taneously, we  find 


Pi  = 


8000   ,    670 


+  -^  =  4168,  and  Qi  = 


8000       670 


3832  pounds. 


Art.  49  253 

If  the  armature  were  not  spinning  (co  =  o),  or  the  car  were  running  on  a  straight 
track  (^  =  o)  then  /c<j12  would  equal  zero,  and  hence  the  reactions  Pi  and  Qi 
would  equal  4000  pounds.  Thus  the  effect  of  the  spin  and  precession  is  to 
increase  one  reaction  and  decrease  the  other  by  670  -^  4  =  168  pounds.  This 
increase  and  decrease  are  called  the  gyrostatic  couple  or  gyrostatic  effect. 
The  force  P2  does  not  depend  on  the  spin  of  the  armature,  only  on  the  radius 
of  the  curve  and  the  velocity  of  the  car.  It  is  often  described  as  the  centrif- 
ugal effect. 

(2)  Fig.  386  represents  a  pair  of  car  wheels  which  we  assume  to  be  rounding 
a  curve.  We  will  determine  the  forces  acting  on  them.  We  assume  that  the 
wheels  are  "coned"  so  that  there  is  true  rolling;  even  if  there  were  slipping  — 
because  of  the  excess  length  of  the  outer  over  the  inner  rail  —  our  results  would 
be  practically  correct.  We  neglect  the  tilt  of  the  track  and  so  regard  the 
precession  as  normal.  Let  W  =  the  weight  of  the  wheels  (including  their 
axle),  M  =  their  mass,  k  =  the  radius  of  gyration  of  wheels,  r  =  their  radius, 
T'  =  velocity  of  the  center  of  gravity,  R  =  the  radius  of  the  curve,  and  /  = 
gage  of  the  track.  Further  let  P  and  Q  =  the  vertical  components  of  the 
pressure  of  the  outer  and  inner  rails  on  the  wheels;  H  =  the  transverse  com- 
ponent of  the  pressure  of  the  outer  rail.  Besides  these  there  are  components 
along  the  rails  with  which  we  are  not  concerned.  According  to  the  last  of 
equations  (i), 

P{R  +  y)+Q(R-  hf)  -WR-Hr  =  MkWyRv, 

and  according  to  the  first  two  of  equations  (2) 

H  =  MVyR  and  P  -f  ()  =  W. 

Solving  these  three  simultaneously  for  P  and  Q  we  get 

_,      W  .  MVh.MkW- 

IRT  _  MVh  _  MkW^ 
^~  2  Rf  Rrf 

The  first  terms  in  these  two  expressions  are  due  to  gravity.  The  second 
terms  are  due  to  centrifugal  action;  they  have  the  same  values  as  if  the  wheels 
were  skidding,  that  is,  they  do  not  depend  on  the  spin  of  the  wheels.  The 
third  terras  are  due  to  gyrostatic  action;  the  components  of  P  and  Q  which 
they  stand  for  constitute  the  so-called  gyrostatic  couple. 

§  3.  Gyrostatic  Reaction.  —  In  general,  any  system  of  forces  can  be 
compounded  into  a  single  force  acting  through  any  desired  point  and  a  couple 
(Art.  9).  Let  us  imagine  all  the  forces  acting  on  a  gyrostat  which  is  precess- 
ing  normally,  to  be  compounded  into  a  force  acting  through  the  mass-center 
of  the  gyrostat  and  a  couple.  Let  these  be  denoted  by  F  and  C  respectively; 
also  let  N  and  n  respectively  =  the  number  of  precessional  and  spinning 
revolutions  per  unit  time,  W  =  the  weight  of  the  gyrostat,  and  k  =  its 
radius  of  gyration  about  the  axis  of  spin. 


2^4  Chap,  xi 

It  follows  from  equations  (2),  that  F  is  directed  from  the  mass-center  Q  to  0 
(Fig.  384)  and 

F  =  MrS^2  =  (p^/g)  r/^Trm\  (3) 

When  0  =  0,  the  torques  of  F  and  C  together,  about  the  x,  y,  and  s  axes, 
must  equal  o,  /coO,  and  o  respectively.  But  F  has  no  torques  about  these 
axes;  hence  C  has  no  torques  about  the  x  and  2  axes,  and  its  torque  about  the 
y  axis  equals  /col].  Therefore  the  plane  of  the  couple  C  is  normal  to  the  per- 
pendicular to  the  spin  and  precession  axes,  and 

C  =  l0i^=    {W/g)  k^  4  TT^wiV.  (4) 

The  sense  of  the  couple  may  be  described  as  follows:  Imagine  a  vector  laid 
ofif  on  the  axis  of  spin  to  represent  the  direction  of  the  spin;  then  the  vector 
representing  the  couple  at  any  instant  is  parallel  to  the  position  which  the 
spin-vector  will  occupy  at  the  end  of  a  quarter  of  the  precession  period  (time 
required  for  one  turn  about  the  axis  of  precession).  See  Fig.  384;  NA  is 
the  spin-vector  and  NB  is  the  couple-vector. 

From  the  stated  facts  in  regard  to  F  and  C,  it  should  be  plain  that  equa- 
tions (i)  are  correct. 

The  gyrostat  exerts  reactions  on  the  bodies  which  exert  forces  upon  it  equal 
and  opposite  to  those  forces  respectively.  Hence  those  reactions  are  equiva- 
lent to  —F  and  —C,  where  —F  and  —  C  denote  a  force  and  a  couple  respec- 
tively equal  and  opposite  to  F  and  C.  Now  F  is  independent  of  the  spin  (see 
equation  i)  but  C  depends  on  it.  Hence  —C  is  called  the  gyrostatic  (part  of 
the)  reaction. 

In  the  examples  of  the  preceding  section  we  determined  the  forces  acting  on 
certain  gyrostats,  and  it  is  easy  to  pick  out  the  gyrostatic  reactions.  Thus, 
in  example  (i)  the  armature  shaft  exerts  downward  forces  of  4168  and  3832 
pounds  on  its  left-  and  right-hand  bearings  as  seen  from  the  rear.  As  already 
pointed  out  each  of  these  pressures  is  the  resultant  of  two  components,  thus 

4000  +168  and  4000  —  168; 

the  second  components  are  the  gyrostat  reaction,  that  is  the  couple  denoted 
by  —  C.  In  example  (2)  the  car  wheels  exert  downward  pressures  equal  to 
P  and  Q.  The  third  components  of  these  reactions  constitute  the  gyrostatic 
reaction  of  the  wheels. 

A  side  (paddle)  wheel  steam  boat  sustains  gyrostatic  reactions  in  certain 
circumstances.  When  such  a  boat  is  turning,  the  (pair  of)  paddle  wheels  and 
shaft  exert  a  gyrostatic  couple  on  the  boat  which  makes  the  boat  heel.  When 
the  boat  is,  say,  travelHng  forward  and  turning  to  starboard,  the  couple  heels  the 
boat  to  port.  Likewise  a  screw-propelled  ship  sustains  a  gyrostatic  couple 
when  she  is  turning;  it  is  due  to  the  precession  of  the  screw  and  shaft  (and 
turbine  too  if  so  equipped).     The  couple  depresses  the  bow  or  stern  depending 


Art.  49  255 

on  the  direction  of  turning  of  the  ship  and  sense  of  rotation  of  the  screw.  It 
has  been  suggested  that  the  gyrostatic  reactions  to  which  (comparatively  frail) 
torpedo-boat  destroyers  are  subject  may  over-tax  their  strength.  The  fact 
is,  these  reactions  are  quite  insignificant  compared  to  other  straining  actions 
which  such  boats  withstand  (see  J.  and  J.  G.  Gray's  "Treatise  on  Dynamics," 
page  531). 

A  flying  machine  is  subjected  to  a  gyrostatic  reaction  of  its  propeller,  shaft 
and  engine  when  turning  or  when  describing  any  curved  path.  When  turning, 
the  reaction  tends  to  raise  or  depress  the  front  of  the  machine,  depending  on  the 
circumstances.  Propellers  being  right-hand  screws  (turning  clockwise  when 
viewed  from  the  rear),  the  front  is  raised  (unless  prevented  by  the  air  man) 
when  he  turns  to  the  left.  When  he  makes  a  dive  the  couple  tends  to  advance 
the  side  of  the  machine  on  the  right-hand  side  of  the  air  man.  The  flight  of 
a  machine  fitted  with  two  screws  which  rotate  in  opposite  directions  is  not  thus 
interfered  with  by  gyrostatic  reactions.  Each  propeller  exerts  a  couple  on 
the  machine  but  the  two  couples  are  always  opposite.  It  has  been  suggested 
that  gyrostatic  reactions  of  propellers  and  motors  may  have  been  the  cause 
of  some  flying-machine  accidents.  However,  a  well-built  machine  can  safely 
withstand  such  reactions  even  under  conditions  of  legitimate  quick  diving 
and  turning.  Thus,  for  a  dive  or  turn  at  the  rate  of  one  revolution  in  20 
seconds,  it  has  been  ascertained*  that  a  loo-horse-power  Gnome  motor  —  speed 
not  stated,  but  probably  about  1 200  revolutions  per  minute  —  exerts  a  gyro- 
static couple  of  140  foot-pounds;  and  the  (suitable)  propeller,  a  couple  of 
184  foot-pounds.  The  forces  involved  in  the  couples  come  upon  the  flying 
machine  at  the  supports  of  the  engine  and  the  propeller  shaft,  f 

*  M.  O'Gorman  in  The  Aeronautical  Journal  for  April,  1913. 

t  For  a  full  discussion  of  the  subject  of  this  article,  consult  Crabtree's  Spinning  Tops  attd 
Gyroscopic  Motion. 


CHAPTER    XII 

TWO   DIMENSIONAL    (PLANE)    MOTION 
50.   Kinematics  of  Plane  Motion 

§  I.  Plane  motion  is  a  motion  in  which  every  point  of  the  moving  body 
remains  at  a  constant  distance  from  a  fixed  plane.  Each  point  of  the  body 
moves  in  a  plane;  that  is,  its  motion  is  uniplanar.  By  plane  of  the  motion  is 
meant  the  plane  in  which  the  mass-center  of  the  body  moves.  The  wheels  of 
a  locomotive  running  on  a  straight  track  have  plane  motion ;  also  a  book  which 
is  slid  about  in  any  way  on  the  top  of  a  table.  A  translation  (Art.  35)  may  or 
may  not  be  a  plane  motion;  a  rotation  about  a  fixed  axis  (Art.  37)  is  always  a 
plane  motion. 

In  a  plane  motion  all  points  of  the  moving  body  which  lie  on  a  perpendic- 
ular to  the  plane  of  the  motion  move  alike,  and  the  motion  of  the  projection 
of  this  line  on  the  plane  of  the  motion  correctly  represents  the  motion  of  all 
the  points.  So  also  the  motion  of  the  projection  of  the  moving  body  upon 
the  plane  of  the  motion  correctly  represents  the  motion  of  the  body  itself. 
Thus  we  have  a  plane  figure  (the  projection  just  mentioned)  moving  in  a 
plane  representing  a  plane  motion  of  a  body;  and  since  the  motion  of  the  plane 
figure  is  uniplanar,  the  motion  of  the  body  is  called  uniplanar.  Hereafter, 
we  will  sometimes  refer  to  the  projection  of  the  body  as  the  body  itself. 

By  angular  displacement  of  a  body  whose  motion  is  plane  is  meant  (as  in 
rotation)  the  angle  described  by  any  line  of  the  body  which  is  in  the  plane  of 
the  motion.     Obviously  all  such  fines  describe  equal  angles  in  the  same  in- 
terval of  time.     As  in  rotations  also,   displacements  are 
regarded  as  positive  or  negative  according  as  they  are  due 
to  counter-clockwise  or  clockwise  turning  of  the  body.     Let 
the  irregular  outline  (Fig.  387)  represent  the  projection  of 
the  moving  body  on  the  plane  of  the  motion,  AB  n.  fixed 
^  ^  fine  of  the  projection,  and  OX  a  fixed  reference  fine;  also 

let  6  denote  the  angle  XOA,  it  being  regarded  as  positive  or  negative  ac- 
cording as  OX,  when  turned  about  0  toward  AB,  turns  counter-clockwise  or 
clockwise.  If  di  and  62  denote  initial  and  final  values  of  6  corresponding  to 
any  motion  of  the  body,  then  the  angular  displacement  =  62  —  Oi  =  AG. 

If  a  body  has  a  plane  motion,  its  angular  velocity  is  the  time-rate  at  which 
its  angular  displacement  occurs,  and  its  angular  acceleration  is  the  time-rate  at 
which  its  angular  velocity  changes.  These  definitions  are  precisely  similar  to 
those  of  the  angular  velocity  and  acceleration  of  a  rotation  about  a  fixed  axis 

256 


Art.  50 


257 


(Art.  37);  hence  the  expressions,  units,  and  rules  of  signs  given  in  that  article 
hold  also  for  any  plane  motion.     The  expressions  are 

0)  =  dd/dt     and     a  =  do)/dt  =  d~d/df, 

CO  and  a  denoting  angular  velocity  and  acceleration  of  the  moving  body  re- 
spectively. 

§  2.  Any  uniplanar  displacement  of  a  body  can  be  accomplished  by  means 
of  a  translation  of  the  body  followed  by  a  rotation,  or  vice  versa.  Thus  let 
AiBiCi  (Fig.  3S8)  be  one  position  of  a  body  ABC,  and  A2B2C2  a  subsequent 
position.  By  means  of  a  translation  the  body  can  be  displaced  so  that  one 
of  its  points  is  put  into  its  final  position;   thus  a  translation  to  A2b'c'  puts  A 


B,  ...- 


Fig.  388 

into  its  final  position.  Then  a  rotation  of  the  body  about  A2  puts  the  body 
into  its  final  position.  Or,  by  means  of  a  rotation  we  can  put  the  body  into 
an  intermediate  position  Aib"c"  so  that  each  Hne  in  it  will  be  parallel  to  its 
final  position  (in  A2B2C2);  and  then  the  body  may  be  put  into  its  final  position 
by  a  translation.  Obviously,  the  translation  and  rotation  might  be  performed 
simultaneously. 

The  point  (or  axis)  of  the  body  about  which  we  imagine  the  rotation  to 
occur  is  called  a  base  point  (or  base  axis).  Fig.  388  also  represents  a  displace- 
ment from  AiBiCi  to  A2B2C2,  accompHshed  with  B  as  base  point.  A  trans- 
lation puts  the  body  into  the  position  B2a"'C",  and  a  suitable  rotation  about 
B2  puts  it  into  the  final  position  B2A2C2.  It  is  clear  that  the  amount  of  the 
translation  component  depends  on  the  base  point;  thus  A1A2  is  the  transla- 
tion for  A  as  base  point,  while  B1B2  is  the  translation  for  B  as  base  point. 
But  the  amount  of  the  rotation  component  does  not  depend  on  the  base 
point;  thus  the  rotation  is  the  angle  ^'.42^2  for  A  as  base  point,  and  it  equals 
the  angle  a"'B2A2  which  is  the  rotation  for  B  as  base  point. 

The  successive  small  displacements  of  ABC  from  AiBiCi  to  A'B'C,  A"B"C", 
etc.,  to  A2B2C2  (Fig.  389)  already  mentioned  (and  which  altogether  approxi- 
mate to  a  continuous  motion  of  ABC  in  which  all  points  of  the  body  move  along 
smooth  curves),  can  each  be  made  by  a  small  simultaneous  translation  and 
rotation.  And  if  we  take  some  one  point  as  base  point  for  all  these  small  dis- 
placements then  we  may  regard  the  motion  as  a  continuous  combined  or 


2S8 


Chap,  xii 


simultaneous  translation  and  rotation,  the  translation  being  like  the  motion 
of  the  base  point  and  the  rotation  being  about  that  point.  In  accordance 
with  this  view,  the  velocity  of  any  point  of  the  moving  body  at  any  particular 
instant  consists  of  two  components,  one  corresponding  to  the  translation  and 
one  to  the  rotation.  Thus  let  A  (Fig.  390)  be  the  chosen  base  point,  v'  =  the 
velocity  of  A  for  the  position  of  the  body  shown,  and  co  =  the  angular  velocity 


Ar'A' 


-.Bg 


Fig.  389 


of  the  body  at  the  instant  under  consideration.  Then  the  first  component  of 
the  velocity  of  any  point  P  equals  v'  and  is  directed  like  v';  the  second  compo- 
nent equals  rco  {r  =  AP)  and  is  directed  at  right  angles  to  AP,  the  sense 
depending  on  the  sense  of  co  (clockwise  or  counter-clockwise) .  Also  the  accelera- 
tion of  any  point  consists  of  two  components,  one  corresponding  to  the  trans- 
lation component  of  the  motion  and  one  to  the  rotation.  Thus  let  a'  be  the 
acceleration  of  the  base  point,  and  a  =  the  angular  acceleration  of  the  body. 
Then  the  first  component  of  the  acceleration  of  any  point  Q  equals  a'  and  is 


Fig.  391 


/ 

8 -ft/sec  ^"^^A 

Fig.  392 


->■  2  ft/sec/sec  ^ 
30.4  fi-Aec/sec  ^""'"^A 


Q  w///m////////////////mMm^'  ^ 


Fig. 


393 


directed  like  a'\  the  second  component  we  describe  by  means  of  two  com- 
ponents, as  in  a  rotation  about  a  fixed  axis  (see  Art.  37),  one  of  which  (the 
normal  component)  is  directed  along  QA  and  the  other  (the  tangential  com- 
ponent) is  at  right  angles  to  QA.  The  normal  component  equals  rur  {r  =  AQ) 
and  is  always  directed  from  <3  to  ^,  toward  the  base  point  or  center  of  the 
rotational  component;  the  tangential  component  equals  m,  and  obviously 
its  sense  depends  on  the  sense  of  the  angular  acceleration. 

For  a  numerical  example  let  us  consider  the  motion  of  the  bar  AB  (Fig.  391) 
the  ends  of  which  slide  along  the  hues  OA  and  OB.  Let  the  length  of  the  bar  --= 
6  feet,  and  the  velocity  and  acceleration  of  ^4  =  6  feet  per  second  and  2  teet 


Art.  50 


259 


per  second  per  second  respectively  (both  toward  the  right)  when  0  =  30 
degrees.  Required  the  velocity  and  acceleration  of  P,  4  feet  from  A.  It  is 
plain  from  the  figure  that  6  cos^  =  x;  hence, 

—  6  sin  0  dd/dt  =  dx/dt,     or     —  6  sin  d'c^  =  v,  (i) 

where  co  =  the  angular  velocity  of  the  bar  and  v  =  velocity  of  A  at  any 
instant.     Differentiating  the  last  equation  with  respect  to  time  we  get 


-  6  (oj  cos  d'dd/dt  +  sin  d'doi/dt)  =  dv/dt,  or 
6  (a;2  cos  6  -\-  aimd)  =  a, 


(2) 


where  a  =  the  angular  acceleration  of  the  bar  and  a  =  the  acceleration  of  A 
Sit  any  instant.  Now  when  6  =  30",  (i)  gives  co  =  —  2  radians  per  second,  and 
(2)  gives  a  =  —  7.6  radians  per  second  per  second.  The  negative  signs  mean 
that  CO  and  a  are  counter-clockwise,  clockwise  having  been  taken  as  positive  for 
6.  Finally,  the  velocity  components  of  P  are  t>  =  6,  and  4  X  w  =  —  8  feet  per 
second  as  shown  in  Fig.  392;  the  acceleration  components  of  P  are  a  =  2, 
4X0;=—  30.4,  and  4  X  co^  =  16  feet  per  second  per  second  as  shown  in 

Fig.  393- 

§  3.  Any  uniplanar  displacement  of  a  body  can  be  accomplished  by  means 
of  a  single  rotation.  Thus  consider  the  displacement  of  ABC  from  the  position 
AiBiCito  A2B2C2  (Fig.394).  The  points  can  be  brought  from  ^1  to  ^2  by  means 


Fig.  394 


Fig.  395 


B^     B,r- ^T — -vB 


I — ^ — 


Fig.  396 


of  a  rotation  of  AB  about  any  point  on  the  perpendicular  bisector  aO  (of  .4 1^2); 
and  B  can  be  brought  from  Bi  to  B2  by  means  of  a  single  rotation  oi  AB  about 
any  point  on  the  perpendicular  bisector  bO  (of  B1B2).  If  the  intersection  of 
the  bisectors  is  taken  for  the  center  of  rotation  of  both  A  and  B,  then  the 
amounts  of  the  rotations  (angles  A1OA2  and  B1OB2)  are  equal;  hence,  the  line 
AB  (and  body  ABC)  can  be  displaced  from  one  position  to  any  other  (uni- 
planar displacement)  by  means  of  a  single  rotation  as  stated. 

In  case  the  two  bisectors  coincide  (Fig.  395),  then  the  angles  Bi  and  B2  are 
equal  and  hence  the  lines  AiBi  and  .42^2  extended  intersect  on  the  bisector  ab 
extended;  this  extension  is  the  center  of  rotation  C  which  would  disi)lace  AB 
from  AiBi  to  A2B2.    In  case  the  bisectors  are  parallel  (Fig.  396)  the  center  of 


26o  Chap.  xi7 

rotation  is  "  at  infinity,"  and  the  displacement  is  a  translation;  thus  a  uniplanar 
translation  may  be  regarded  as  a  rotation  about  a  center  at  infinity. 

The  actual  continuous  motion  of  AB  from  one  position  AiBi  to  an- 
other A2B2  (in  which  A  and  B  describe  smooth  curves)  can  be  closely  duplicated 
by  a  succession  of  rotations  of  AB  from  AiBi  (Fig.  389)  into  successive  inter- 
mediate positions  A'B',  A"B",  etc.,  until  .42^2  is  reached.  Each  small  rota- 
tion is  made  about  a  definite  center  0',  0",  etc.  (not  shown).  The  closer  these 
intermediate  positions  are  taken  (and  the  more  numerous  and  closer  the  centers 
of  rotation  0',  0" ,  etc.)  the  more  nearly  do  the  successive  rotations  reproduce 
the  actual  continuous  motion.  "In  the  limit,"  the  actual  motion  is  repro- 
duced by  the  rotations,  the  centers  of  rotation  forming  a  continuous  line. 
Thus  we  may  regard  any  uniplanar  motion  of  a  body  as  consisting  of  a  con- 
tinuous rotation  about  a  center  which,  in  general,  is  continuously  moving. 
The  position  of  the  center  0  about  which  the  moving  body  is  rotating  at  any 
instant  is  called  the  instantaneous  center  of  the  motion  for  the  particular  instant 
or  position  (of  the  body)  under  consideration,  and  the  line  through  that  center 
and  perpendicular  to  the  plane  of  the  motion  is  called  the  instantaneous  axis 
of  the  motion  for  that  instant. 

In  general,  the  instantaneous  center  moves  about  in  the  body  and  in  space. 
Its  path  in  the  body  is  called  body  centrode;  its  path  in  space  the  space  cen- 
trode.  Thus,  in  the  case  of  a  wheel  rolling  on  a  plane,  the  instantaneous  center 
at  any  instant  is  the  point  of  contact  between  the  wheel  and  plane;  the 
successive  instantaneous  centers  on  the  wheel  trace  or  mark  out  the  circum- 
ference and  this  line  is  the  body  centrode;  the  successive  instantaneous  centers 
in  space  trace  or  mark  out  the  track  and  this  line  is  the  space  centrode.  It 
can  be  shown  that  any  plane  motion  may  be  regarded  as  a  rolling  of  the  body 
centrode  on  the  space  centrode. 

Now  in  a  rotation  about  a  fixed  axis  the  velocities  of  all  points  of  the  body 
are  proportional  to  the  distances  of  the  points  from  the  axis  of  rotation,  and 
the  velocities  are  respectively  normal  to  the  perpendiculars  from  the  points 
to  the  axis  (Art.  37);  the  velocity  of  any  particular  point  is  given  by  v  =  rw, 
where  v  =  the  velocity  of  the  point,  r  =  the  distance  of  the  point  from  the 
axis,  and  co  =  the  angular  velocity  of  the  body.  So  too,  in  the  case  of  a  uni- 
planar motion,  the  velocities  of  all  points  of  the  body  at  any  particular  instant 
are  proportional  to  the  distances  of  the  points  from  the  instantaneous  axis 
(corresponding  to  that  instant);  the  velocities  are  respectively  normal  to  the 
perpendiculars  from  the  points  to  the  instantaneous  axis;  and  the  velocity  v 
of  any  particular  point  is  given  by  z^  =  rw,  where  r  =  the  distance  from  the 
point  to  the  axis  and  co  =  the  angular  velocity  of  the  body. 

By  means  of  the  foregoing  velocity  relations,  we  can  locate  the  instantaneous 
center  for  any  given  position  of  the  moving  body  if  the  directions  of  the  veloci- 
ties of  two  of  its  points  are  given;  and  then  if  the  value  of  one  velocity  is  given 
we  can  compute  the  angular  velocity  of  the  body  and  the  velocity  of  any  other 
point.     For  an  example  we  will  consider  the  connecting  rod  of  an  engine  {BC, 


Art.  51  261 

Fig.  397),  in  the  position  shown,  the  speed  being  100  revolutions  per  minute. 
Since  the  velocity  of  the  point  B  of  the  rod  is  along  the  tangent  to  the  crank- 
pin  circle  at  B,  the  instantaneous  center  of  the 
connecting  rod  is  on  the  normal  to  the  tangent 
at  B,  that  is  on  AB  or.  its  extension;  and  since 
the  velocity  of  the  point  C  of  the  rod  is  along 
AC,  the  instantaneous  center  is  on  the  normal     C'  \  i 

to  AC.     Hence  the  instantaneous  center  is  at  the  -pic 

intersection  O.     Now  velocity  oi  B  —  2  t  Y.  AB 

(to  scale)  X  100  =  2000  feet  per  minute;  hence,  the  angular  velocity  of  the 
rod  =  2000  -^  OB  (to  scale)  =  185  radians  per  minute.  The  velocity  of 
C  =  OC  (to  scale)  X  185  =1110  feet  per  minute. 

51.   Kinetics  of  Plane  Motion 

§  I.  General  Principles.  —  From  the  principle  of  the  motion  of  the 
mass-center  (Art.  34)  we  may  write  at  once 

^F,  =  Ma,,    HFy  =  May,     and    SF,  =  o;  (i) 

where  SF^,  'ZFy,  and  HF^  =  the  algebraic  sums  of  the  components  of  the  ex- 
ternal forces  acting  on  the  body  along  three  rectangular  lines,  the  third  one 
being  at  right  angles  to  the  plane  of  the  motion,  a,  and  ay  respectively  =  the 
X  and  y  components  of  the  acceleration  of  the  mass-center,  and  M  =  the  mass 
of  the  body.  In  addition  to  the  above,  we  have  another  simple  relation 
(established  later), 

f  =  la  =  Mk'-a  (2) 

where  T  denotes  the  torque  of  all  the  external  forces  about  the  line  through 
the  mass-center  and  perpendicular  to  the  plane  of  the  motion,  J  =  the  moment 
of  inertia  of  the  body  about  the  line  just  mentioned,  ~k  =  the  radius  of 
gyration  of  the  body  about  that  hne,  and  a  =  the  angular  acceleration  of  the 
moving  body.  Systematic  units  (Art.  31)  must  be  used  in  equations  (i) 
and  (2).  But  we  may  substitute  W/g  for  M  (where  W  =  the  weight  of  the 
body  and  g  =  the  acceleration  due  to  gravity)  and  then  use  any  convenient 
units  for  force  (and  weight),  length,  and  time. 

To  derive  equation  (2),  let  Fig.  398  represent  the  moving  body,  C  be  the 
mass-center,  a  =  the  acceleration  of  C,  00  and  a  =  the  angular  velocity  and 
acceleration  respectively  of  the  body.  Further,  let  Pi,  P2,  etc.,  be  particles  of 
the  body;  mi,  m^,  etc.,  =  their  masses;  Vi,  r^,  etc.,  =  their  distances  from  the 
line  through  C  and  perpendicular  to  the  plane  of  the  motion;  and  Ri,  R2, 
etc.,  =  the  resultants  respectively  of  all  the  forces  acting  on  Pi,  P^,  etc.  We 
will  regard  the  motion  as  consisting  of  a  translation  Uke  the  motion  of  C  and  a 
rotation  about  the  "  base  axis  "  through  C.  Then  the  acceleration  of  Pi  can 
be  regarded  as  consisting  of  three  components,  a,  ria,  and  rico-  as  indicated; 
likewise  the  acceleration  of  P2  can  be  regarded  as  consisting  of  three  com- 


262 


Chap,  xn 


ponents,  a,  r^a,  riw-;  etc.  Therefore,  the  resultant  Rx  consists  of  three  com- 
ponents wia,  wiria,  and  Wirio;-  directed  like  the  corresponding  accelerations; 
similarly,  the  resultant  Ri  consists  of  three  components  n^fl,  nhT^a,  and  m^r-ior 
directed  like  the  corresponding  accelerations;  etc.  Now  the  torque  of  all  the 
forces  acting  on  Pi  =  the  torque  of  the  (three)  components  of  Ri,  similarly, 
the  torque  of  all  the  forces  acting  on  P2  =  the  torque  of  the  (three)  com- 
ponents of  R2;    etc.    Hence,  the  torque  of  all  the  forces  acting  on  all  the 


r.a 


Fig.  3g8 


Fig.  399 


particles  (external  and  internal  forces  acting  on  the  body)  =  the  torque  of 
the  components  (as  nia,  mra,  and  mm^)  of  all  the  resultants  Ri,  R2,  etc.  Since 
the  internal  forces  occur  in  pairs  of  equal,  opposite,  and  colinear  forces,  they 
contribute  nothing  collectively  to  the  first  torque  just  mentioned.  It  is  plain 
from  the  figure  that  the  normal  components  mirioj-,  nhVoj^-,  etc.,  have  no 
torque  about  the  (base)  axis.  Since  the  resultant  of  the  components  WiO, 
fn^,  etc.,  passes  through  the  mass-center  (Art.  35),  they  have  no  torque  about 
the  axis.     The  torque  of  the  remaining  set  of  components  is 


Wi^iari  +  miTiari  + 


=  al 


(see  Art.  36).     Hence,  we  have  T  =  /a,  or  equation  (2). 

Three  Special  Cases.  —  (i)  When  the  velocity  of  the  mass-center  is  con- 
stant in  amount  and  direction  {a  =  o),  the  torque  of  the  external  forces  about 
any  line  perpendicular  to  the  plane  of  the  motion  equals  la.  iii)  When  the 
angular  velocity  is  constant  (a  =  o),  the  torque  of  the  external  forces  about  a 
line  through  the  mass-center  and  perpendicular  to  the  plane  of  the  motion 
equals  zero.  {Hi)  When  a  and  a  =  o,  the  torque  of  the  external  forces  about 
any  line  perpendicular  to  the  plane  of  the  motion  equals  zero. 

Examples.  —  i.  Required  the  value  of  P  for  starting  a  wheel  (Fig.  399a) 
or  stopping  it  (Fig.  399b).  The  figures  show  the  wheel  rolling  toward  the 
right.  In  the  two  figures  respectively,  the  angular  accelerations  are  clock- 
wise and  counter-clockwise;  hence  the  friction  F  on  the  wheels  act  as  shown, 
and  F  =  \D  =  Mk~a.  And  since  the  accelerations  a  of  the  mass-center  are 
toward  the  right  and  left  respectively,  P  -  F  =  Ma  iox  each  figure.  Also 
a  =  ra.  From]  these  equations,  it  follows  that  P  =  M  (i  +  k'^/r'^)  a.  Thus 
the  "effective  inertia  "  of  a  rolling  wheel  is  i  +  k'^/r'^  times  its  inertia  when 
skidding,  for  in  the  latter  case  P  —  Ma. 


Art.  51  263 

2.  It  is  required  to  discuss  the  rolling  of  a  homogeneous  cylinder  on  an 
inclined  plane.  Let  the  weight  of  the  cylinder  =  200  pounds,  the  diameter 
of  its  bases  =  3  feet,  and  the  inclination  of  the  plane  =  25  degrees.  Further, 
we  assume  that  the  cylinder  and  plane  do  not  distort  each  other,  so  that  there 
is  only  line-contact  between  them  and  no  "rolling  resistance"  (Art.  52);  also 
that  the  surfaces  in  contact  are  sufficiently  rough  to  prevent  slipping  so  that 
the  roUing  is  perfect.     There  are  only  two  external  forces  acting  on  the  rolling 

Wj ^ 

Fig.  400  Fig.  401 

cylinder,  its  own  weight  and  the  reaction  of  the  plane,  but  the  latter  is  repre- 
sented by  two  components,  N  and  F,  in  Fig.  400.  Since  the  mass-center 
moves  in  a  line  parallel  to  the  incline,  ax  =  a,  and  ay  =  o;  hence  equations 
(i)  become 

200  sin  25°  —  F  =  (200  -^  32.2)  a, 
N  —  200  cos  25°  =  o,   and  0  =  0. 

The  second  equation  shows  that  iV  =  181  pounds.  The  first  equation  con- 
tains two  unknowns  (F  and  a)  and  does  not  furnish  the  value  of  either  of  them; 

so  we  resort  to  equation  (2).  Since  ^  =  |  1.5-  =  1.125  (see  Art.  36),  equa- 
tion (2)  becomes 

F  X  1.5  =  (200  -^  32.2)  1. 125  X  oc. 

Now  we  have  two  equations  but  three  unknowns,  and  so  we  need  an  ad- 
ditional equation;  this  is  given  by  the  (simple)  relation  between  a  and  a. 
Since  there  is  no  slipping,  the  displacement  s  of  the  mass-center  in  any  interval 
of  time  and  the  angular  displacement  6  of  the  cyUnder  for  that  interval  are 
related  thus:  .j  =  1.5  0  (6  in  radians  and  s  in  feet);  hence  d^s/df^  =  i-5  d^O/df^, 
or  a  =  1.5  a.  Substituting  1.5  a  for  a  in  the  first  equation  and  then  solving 
simultaneously  V\^ith  the  fourth,  we  find  that  a  =  6.05  radians  per  second  per 
second  (o  =  9.07  feet  per  second  per  second)  and  F  =  28. 2  pounds. 

3.  It  is  required  to  discuss  the  forces  acting  on  a  rolling  wheel  whose  center 
of  gravity  is  not  in  the  axis  of  the  wheel,  the  speed  of  rolling  being  maintained 
uniform  by  a  suitable  horizontal  force  P  (Fig.  401).  Let  W  =  weight  of  the 
wheel,  r  —  radius,  and  c  =  the  distance  from  its  center  A  to  the  center  of 
gravity  C;  further  let  6  =  the  angle  between  AC  and  the  horizontal  in  the 
position  of  the  wheel  under  consideration.  There  are  three  forces  acting  on 
the  wheel,  P,  W,  and  the  reaction  of  the  roadway  (represented  for  convenience 
by  two  components  A^  and  F).     Equations  (i)  become 

P-  F  =  (Pf /g) ~a,    and    N  -  W  =  {W/g) ay. 


264  ^^p-  ^^ 

Since  the  angular  velocity  is  constant,  a  =  o,  and  equation  (2)  becomes 
F  (r  +  c  sin  6)  —  Nc  cos  d  —  Pc  sin  d  =  o. 

These  equations  contain  five  unknowns  {P,  F,  N,  Ux,  and  ay),  and  so  we 
need  other  equations.  Obviously  the  relations  between  a^,  ay,  and  d  furnish 
the  additional  equations.  To  determine  these  let  us  regard  the  rolling  as  con- 
sisting of  a  translation  with  A  as  base  point  and  a  rotation  about  A .  Then 
since  A  moves  uniformly,  the  acceleration  of  the  translational  component  =  o; 
and  there  being  no  angular  acceleration,  the  acceleration  of  the  rotational 
component  of  the  motion  of  C  is  wholly  radial  (along  CA)  and  equals  cor. 
Hence  a  equals  cw^  and  is  directed  from  C  to  .4 ;   and 

'     a^  =  ceo- cos  0,     and     ay  =  —  coi"^  sinQ. 

Substituting  these  values  of  a^  and  ay  in  the  first  two  equations,  and  solv- 
ing them  simultaneously  with  the  third  we  find  that 

For  CO  we  may  write  2  ttw,  where  n  =  the  number  of  turns  of  the  wheel  per 
unit  time. 

It  follows  from  the  foregoing  results  that  P  and  F  are  always  opposite; 
that  P  and  F  act  as  shown  whenever  the  center  of  gravity  C  is  on  the  left  of 
the  vertical  through  the  center  A  id  between  —  90°  and  +  90°) ;  that  P  and 
F  act  opposite  to  the  directions  indicated  in  the  figure  when  C  is  on  the  right 
of  the  vertical  through  A ;  that  N  always  acts  upward  unless  cor  sin  6  is 
greater  than  g;  that  the  greatest  value  of  N  obtains  when  C  is  vertically  be- 
low A  {d  -=-  90°)  and  then  N  =  W  (i  +  coi'^/g).  This  excess  TFccoVg  over 
W  in  the  value  of  N  is  called  "  hammer  blow  "  in  locomotive  parlance,  but 
the  hammer  blow  of  a  locomotive  driving  wheel  depends  also  upon  the  side 
rods  attached  to  the  wheel  (see  Art.  35). 

Independence  of  Translation  and  Rotation.  —  Referring  to  equations  (i), 
page  261,  it  will  be  noted  that  they  contain  no  term  depending  on  the  rota- 
tion of  the  body  about  the  mass-center;  therefore,  they  show  that  the  motion 
of  the  mass-center  is  entirely  independent  of  the  rotation  about  that  point. 
And  as  already  pointed  out  (Art.  34),  the  acceleration  of  the  mass-center  is 
the  same  as  though  the  entire  body  were  concentrated  at  the  mass-center  and 
all  the  external  forces  were  applied  at  that  point  parallel  to  their  actual  lines 
of  action.  Equation  (2)  contains  no  term  depending  on  the  motion  of  the 
mass-center;  therefore,  the  rotation  of  the  body  about  the  mass-center  is 
independent  of  any  motion  of  the  mass-center  itself.  And  on  comparing 
equations  (2)  with  the  equation  of  motion  for  rotations  about  fixed  axes 
(Art.  37),  it  becomes  plain  that  the  external  forces  produce  rotation  about  a 
free  (moving)  axis  through  the  mass-center  as  though  that  axis  were  fixed. 
Thus  we  have  complete  independence  of  translation  (of  mass-center)  and 
rotation  (about  mass-center). 


Art.  51 


265 


To  illustrate  we  will  apply  the  principle  of  independence  to  explain  center  of 
percussion;   Art.  48  includes  an  explanation  based  on  other  principles.     Let 
AB  (Fig.  402)  be  a  prismatic  bar  lying  on  a  horizontal  surface,  and  C  its  center 
of  gravity.     Now  imagine  the  bar  to  be  struck  a  blow  in  the  line 
F.     The  only  other  forces  acting  on  the  bar  are  gravity  and  the        ra 
supporting  force  of  the  surface;  these  produce  no  appreciable  \ 

effect  on  the  motion  during  the  blow.     The  motion  produced,  [ 

therefore,  consists  of  a  translation  as  though  the  blow  F  acted  -k- 
through  the  mass-center,  and  a  rotation  about  the  mass-center  __^ 
as  though  the  mass-center  were  fixed.  Any  point  beyond  C 
gets  a  velocity  toward  the  right  due  to  the  translation,  and  a 
velocity  toward  the  left  due  to  the  rotation.  For  some  par-  Fig.  402 
ticular  point  these  two  velocities  are  equal  and  opposite,  and 
hence  if  the  bar  were  pivoted  there,  the  pivot  would  feel  no  pressure  from  the 
bar  during  the  blow.  For  such  a  point,  G  is  the  center  of  percussion.  Let  us 
now  find  where  this  pivot  point  is.  For  that  purpose  let  M  =  mass  of  the 
bar,  k  =  its  radius  of  gyration  about  the  line  through  C  perpendicular  to 
the  supporting  surface,/  =  the  arm  of  the  blow  F  about  the  mass-center,  R  be 
the  pivot  point,  r  =  its  distance  from  C,  a  =  the  average  acceleration  of  the 
mass-center,  a  =  the  average  angular  acceleration  of  the  body  during  the  blow, 
and  M  =  the  duration  of  the  blow.  The  velocities  of  R  due  to  the  translation 
and  rotation  respectively  equal  a  At  and  raAt.     Now 

a  =  F/M    and    a  =  Ff/Mk^l 

therefore,  for  the  pivot  point  we  have 

(f/M)  At  =  r{Ff/Mt)  At,     or    fr  =  t. 


r2 


That  is,  r  =  yfe  //.  For  a  given  pivot  the  distance  of  the  center  of  percussion 
from  the  center  of  gravity  is/  =  ^  /r,  which  agrees  with  the  result  reached  in 
Art.  48. 

Kinetic  Energy  of  a  Body  with  Plane  Motion.  —  Let  M  =  the  mass  of  the 
body,  W  =  its  weight,  7  =  its  moment  of  inertia  about  a  line  through  the 
mass-center  perpendicular  to  the  plane  of  the  motion,  k  =  its  radius  of  gyra- 
tion about  the  same  Une,  v  =  the  velocity  of  the  mass-center,  and  co  =  the 
angular  velocity  of  the  body.     Then  the  kinetic  energy  of  the  body  equals 


im'  +  hr<^'=h  (w/gw  +  \  {wig)k  i^\  (I) 

The  latter  is  the  more  convenient  form  generally  for  use  in  a  numerical  case. 
If  g  is  taken  as  32.2  (feet  per  second  per  second),  then  the  foot  and  second 
should  be  adhered  to  as  units  of  length  and  time;  co  should  be  expressed  in 
radians  per  unit  time.  If  W  be  ex-pressed  in  pounds,  tons,  etc.,  then  the  result 
will  be  in  foot-pounds,  foot-tons,  etc. 


266 


Chap,  xn 


Fig.  403 


The  first  term  of  (i)  equals  the  kinetic  energy  which  the  body  would  have 
if  its  motion  were  one  of  translation  with  velocity  equal  to  v;  and  the  second 
term  equals  the  kinetic  energy  which  it  would  have  if  its  motion  were  one  of  rota- 
tion about  a  fixed  axis  through  the  mass-center  and  perpendicular  to  the  plane 

of  the  motion.  Hence  the  kinetic  energy  of  a 
body  with  any  plane  motion  may  be  regarded 
as  consisting  of  two  parts;  they  are  called  trans- 
lational  and  rotational. 

The  following  is  a  derivation  of  the  preceding 
formula  after  the  view  that  a  plane  motion  is  a 
combined  translation  and  a  rotation  (Art.  50,  §  2). 
Let  Fig.  403  represent  the  moving  body,  0  its 
mass-center,  and  P  any  other  point  of  the  body. 
Also  let  r  =  the  distance  of  P  from  the  fine 
through  0  perpendicular  to  the  plane  of  the  motion,  and  v  =  the  velocity  of  P. 
Then  v  is  the  resultant  of  v  and  rw  as  indicated.  The  angle  QPS  =  90  — 
(iS  —  0),  where  /3  and  6  are  the  angles  which  v  and  OP  respectively  make  with 
the  X  axis.     Therefore 

^)2  —  ^2  _|_  ^2^2  _|_  2  ^/-co  sin  (^  —  6), 
and  the  kinetic  energy  of  the  entire  body  (2  ^  mv~)  equals 

I  v'^^m  +  I  w-Swr-  +  2  I'co  (sin  /SSwr  cos  6  —  cos  ^  Hmr  sin  9). 

Now  r  cos  6  and  r  sin  0,  respectively,  equal  the  x  and  y  coordinate  of  P.  Hence 
Swr  COS0  =  2wx  =  xHm  (see  page  158),  x  denoting  the  x  coordinate  of  the 
mass-center;  and  since  x  =  o,  '^mr  cos  9  =  o.  Similarly,  Swr  sin  9  ^  o. 
Hence  the  foregoing  expression  for  the  kinetic  energy  reduces  to 

The  follo\Adng  is  a  derivation  based  on  the  view  that  any  plane  motion  con- 
sists of  a  succession  of  instantaneous  rotations  (Art.  50,  §  3).  Let  /  ^  the 
moment  of  inertia  of  the  body  about  that  Hne  which  is  the  (instantaneous) 
axis  of  rotation  at  the  instant  in  question,  d  =  the  distance  from  that  axis  to 
the  mass-center,  p  =  the  distance  of  any  point  P  of  the  body  from  the  axis, 
z^  =  velocity  of  P  (as  before),  and  co  =  angular  velocity  of  the  body. 
Then  v  =  pw,  and  the  kinetic  energy  of  the  body  is 


SI  mv^  =  \  w-2wp2  =  i  /aj2 


(2) 


This  is  a  much  simpler  ex-pression  than  (i)  but  not  so  convenient  to  use  gener- 
ally, because  /  refers  to  an  axis  not  fixed  in  the  body.  It  remains  to  reduce 
(2)  to  (i).  According  to  the  parallel  axis  theorem  (Art.  36,  §  2),  /  =  /  + 
Md-\  hence 


|/a;2 


1  /co2  +  i  M{d  •  a))2  =  1  /co2  +  i  .1/^' 


Art.  51 


267 


For  an  example  we  will  compute  the  kinetic  energy  of  a  solid  cylinder 
rolling  on  a  plane  surface.  Let  W  =  weight  of  cylinder,  D  =  its  diameter, 
and  n  =  number  of  turns  of  the  cylinder  per  unit  time.  Then  M  =  W/g, 
V  =  tDh,  7=1  iW/g)D~  (see  Art.  36),  and  co  =-  2  im.  Hence  the  kinetic 
energy  of  the  cylinder  equals 

i  (W/g)TrWhl'  +  i  (W/g)TrWhi\ 

Thus  it  appears  that  two-thirds  of  the  energy  is  translational  and  one-third 
is  rotational. 

§  2.  Dynamics  of  a  Simple  Moving  Vehicle.  ^  Let  W  =  weight  of 
the  body  of  the  vehicle  and  its  load,  if  any;  w  =  the  weight  of  each  wheel  (in- 
cluding one-half  of  the  axle  if  the  wheels  are  rigidly  mounted  on  their  axles) ; 
k  =  radius  of  gyration  of  wheel  (with  one-half  of  axle  in  case  mentioned); 
r  =  radius  of  wheel;  n  =  number  of  wheels;  and  v  =  velocity  of  the  vehicle. 
The  kinetic  energy  of  each  wheel  is 

\  (w/gy  +  h  (Wg)k'  (v/ry  =  i  (w/g)  (i  +  k^r^y. 
Hence  the  kinetic  energy  of  the  entire  vehicle  is 

1 VW      nw 

47"^ 


g 


-^)} 


Comparing  this  expression  with  that  for  the  kinetic  energy  of  a  body  with  a 
motion  of  translation,  we  see  that  the  motion  of  the  entire  vehicle  may  be 
regarded  as  one  of  translation  provided  that  the  weight  of  the  vehicle  is  taken 
equal  to  W  +  nw  (i  -\-  k^/r-).  For  modern  freight  cars  r  =  16.5  inches  and 
k  =  9.5  inches  (about);  hence  k-/r^  =  0.35.  Therefore  the  "effective  in- 
ertia "  of  the  wheels  when  rolling  is  about  one-third  greater  than  when  at  rest 
or  skidding. 

Height  of  Draw  Bar.  —  Fig.  404  represents  a  vehicle,  as  a  railroad  car,  being 
dragged  on  a  level  track  by  a  pull  P.     The  other  external  forces  acting  on  the 


W 


/  N 


•77777?7;77T^'777777777777777m777777777777/l^. 


Fig.  404 


H    ^  ^- 


■ra^ 


-tr 


■h 


Fig.  406 


"-5^5^ 


\ 


Q 


car  are  gravity  iyV  -f-  nw)  and  the  reactions  of  the  rails  on  the  wheels  (each 
represented  by  its  horizontal  and  vertical  components).  In  Fig.  405  there  are 
represented  all  the  external  forces  acting  on  one  wheel,  in  Fig.  406  those 
acting  on  the  car  body.  The  pressures  between  axles  and  bearings  are  repre- 
sented by  their  horizontal  and  vertical  components;  axle  friction  is  disregarded. 
Let  a  =  the  acceleration  of   the  car;    then  the  angular  acceleration  of  the 


268  Chap,  xii 

wheels  =  a/r.     Consideration  of  the  forces  on  the  wheel,  equation  (2),  page 
261,  shows  that 

Fr  =  -h?- ,     or     F  =  — -a. 
g      r  g  r^ 

We  have  also  (according  to  equations  i,  page  261) 

e-.  =  ^a,  or  e  =  f (.  +  !>■ 

Consideration  of  the  forces  acting  on  the  car  body  shows  that  P  —  nQ  = 
(W/g)a,  or 


la. 


When  applied  high  up  on  the  car,  P  tends  to  raise  the  rear  end,  decreasing 
the  rear  vertical  axle  pressures  and  increasing  the  forward  vertical  axle  pres- 
sures. When  applied  low,  P  produces  the  opposite  effect.  Obviously,  when 
applied  in  some  certain  line,  P  has  no  such  effect  on  the  vertical  axle  pressures. 
We  will  now  locate  that  line;  let  h  =  its  height  above  the  plane  of  the  axes 
of  the  axles,  and  H  =  the  height  of  the  center  of  gravity  of  the  car  body 
and  its  load  above  that  plane.  When  the  car  is  at  rest  {P  and  Q  =  o),  the 
(vertical)  pressures  of  the  axles  on  the  car  body  take  on  certain  values. 
If,  when  P  (and  nQ)  act  on  the  car  body,  their  resultant  acts  through  the 
center  of  gravity,  then  those  forces  do  not  tend  to  rotate  the  car  body  and 
do  not  affect  vertical  pressures  of  or  on  the  axles  already  mentioned.  Thus, 
to  provide  against  extra  loading  or  unloading  of  axles  by  P  (draw-bar  effect), 
the  moments  of  P  and  nQ  about  the  transverse  horizontal  line  through  the 
center  of  gravity  of  the  car  body  (and  load)  should  balance.  That  is,  we 
should  have  P{H  -  h)  =  nQH,' or 

H 


h  = 


+  (nw/W)  (i  +  k^/r^) 


52,   Rolling  Resistance 

§  I.  Rollers.  —  In  the  present  connection  a  roller  is  taken  to  differ  from 
a  wheel  (of  a  vehicle)  in  that  the  latter  sustains  its  load  indirectly  through  its 
axle,  while  the  former  has  no  axle  but  takes  its  load  directly.  When  a  roller 
(or  wheel)  is  made  to  roll,  it  experiences  more  or  less  resistance  from  the 
track  (or  roadway)  upon  which  it  rolls.  Obviously  the  amount  of  this  resist- 
ance depends  in  large  part  on  the  nature  of  the  surfaces  in  contact  and  on  the 
amount  of  the  pressure  between  them.  In  the  case  of  an  inelastic  roadway 
{A,  Fig.  407)  the  roller  leaves  a  rut,  and  there  is  a  continual  expenditure  of 
energy  in  thus  (permanently)  deforming  the  track  as  well  as  against  friction 
due  to  actual  rubbing  between  roller  and  track.  In  the  case  of  an  elastic 
roadway  (B,  Fig.  407)  also,  there  is  rubbing  between  the  roller  and  the 
deforming  and  recovering  portions  of  the  track  and  consequently  friction 


Art.  52 


269 


loss.*   In  any  case  there  is  expenditure  of  energy  against  the  (internal)  friction 
in  portions  of   the   roller   and  track   which  are  deforming  or  recovering.! 

Let  R  =  the  resultant  reaction  of  the  track 
on  the  roller.  Obviously  the  point  of  ap- 
plication of  R  is  on  the  surface  (or  arc)  of 
contact  between  wheel  and  roadway;  and  it 
will  be  shown  presently  that  this  point  is  in 
front  of  the  vertical  diameter  of  the  roller, 
the  roadway  supposed  to  be  horizontal.  The 
distance  from  this  point  to  the  diameter  is  called  the  coefficient  of  rolling  re- 
sistance; we  will  denote  it  by  c,  and  express  numerical  values  of  the  coefficient 
in  inches.  Obviously  the  coefficient  of  rolling  resistance  depends  on  the 
nature  of  the  wheel  and  roadway,  and  is  greater  for  yielding  surfaces  than  for 
rigid  ones.  It  would  seem  that  the  coefficient  depends  on  the  load  but  in 
certain  cases  at  least  the  coefficient  is  not  influenced  much  by  it.  The 
coefficient  is  claimed  to  be  independent  of  the  radius  of  the  roller;  also 
that  it  varies  as  the  square  root  of  the  radius.  The  precise  way  in  which  the 
coefficient  varies  with  the  conditions  named  has  not  been  established.  Below 
we  give  some  of  the  meager  ex-perimental  data  relating  to  the  matter. 

Coulomb  seems  to  have  made  the  first  ex-periments  to  determine  coefficients 
of  rolUng  resistance.     The  following  are  his  results  for 

Lignum  Vit.e  Rollers  on  Oak  "Pieces" 


Load. 

Diameter  =  2.18  ins. 

Diameter  =  6.55  ins. 

220.5  lbs. 
1102.7 
2205. 

c  =  0.0174  ins. 

.0205 

.0196 

c  =  0.0196  ins. 
.0197 
.0196 

For  the  circumstances  of  these  ex-periments,  it  appears  that  the  coefficient 
does  not  vary  much  with  the  load  or  the  diameter  of  the  roller.  For  elm 
rollers  6.55  and  13. 11  inches  in  diameter  on  oak  pieces.  Coulomb  found  c  = 
0.0327  inches-l 

The  following  also  are  quoted  from  Morin's  Mechanics: 


Oak  Rollers  (diameter  =  7.87  inches)  on  Poplar  Strips 

Width  of  strips. 

Load. 

Coefficient  c. 

0.97  ins. 
3-94 

409  lbs. 

400 

0.00637  ins. 
.00287 

*  See  Phil.  Trans.  Roy.  Soc,  Vol.  166,  Part  i  (1876),  for  experiments  on  rollers  rolling  on 
a  rubber  roadway,  by  Prof.  Osborne  Reynolds. 

t  See  Jour,  and  Trans.  Soc.  of  Engrs.  (London),  Vol.  3,  p.  180  (1912)  for  an  analysis  of  this 
element  of  rollinp;  resistance,  by  Prof.  Herbert  Chatley. 

X  From  English  translation  by  Joseph  Bennett  of  Morin's  Mechanics,  p.  339. 


270 


Chap,  xii 


In  these  experiments,  increasing  the  length  of  bearing  from  0.97  to  2.94 
(about  triple)  more  than  halved  the  coefficient.  Thus  it  appears  that  the 
coefficient  depends  on  the  loading  per  unit  length  of  contact  between  roller 
and  roadway.  But  the  coefficient  probably  does  not  decrease  indefinitely 
with  increase  of  length  of  contact. 

For  some  conditions  the  coefficient  seems  to  vary  as  the  square  root  of  the 
radius  of  the  roller,  that  is 

c  =  </)  V7, 

where  <f>  is  another  coefficient  and  r  =  radius  of  the  roller.  Dupuit  gives  the 
following  average  values: 

Wood  on  wood 0  =  0.0069 

Iron  on  moist  wood .0063 

Iron  on  iron .0044 

Wheel  on  macadam .19 

For  the  conditions  of  his  experiments,*  Prof.  C.  L.  Crandall  takes  the  co- 
efficient of  rolling  resistance  as  proportional  to  the  square  root  of  the  radius, 
that  is  c  =  0  V  r.  Roller  plates  used  were  i|  inches  thick;  rollers  i,  2,  3  and 
4  inches  in  diameter,  all  i|  inches  long  except  the  first  whose  length  was  i 
inch.  Plates  and  rollers  were  used  as  they  came  from  the  plane  and  lathe; 
were  not  polished  or  filed.  Loads  varied  from  350  to  2500  pounds  per  linear 
inch  in  contact.  The  coefficient  did  not  seem  to  vary  much  with  load;  with 
materials  it  varied  as  follows: 

Cast  iron 0  =  0.0063 

Wrought  iron .0120 

Steel -0073 

These  values  refer  to  cast-iron  plates;   for  wrought-iron  plates  they  should 

be  increased  about  13  per  cent,  and  for  steel  plates  they  should  be  decreased 

by  that  amount. 

Fig.  408  represents  in  principle  the  device  used  by  Coulomb  to  determine 

the  coefficient  of  rolling  resistance.  W  =  weight  of  roller,  Wi  and  W2  = 
weights  of  suspended  bodies  as  shown.  By  adjusting  the 
difference  between  Wi  and  W2  the  roller  was  made  to  roll 
quite  uniformly.  When  rolling  at  constant  speed,  the  reaction 
R  of  the  track  on  the  roller  is  vertical,  and  R  =  W  -\-  Wi  +  W^- 
Also  there  is  no  resultant  torque  on  the  roller;  hence  the 
moment  of  R  must  be  counter-clockwise  (in  this  illustration), 
and  so  the  point  of  application  of  R  is  in  front  of  the  vertical 
diameter  of  the  roller  (as  stated).  It  follows  that  (IF2  —  W\) 
/  =  Re  =  (W  +  Wi  -hW2)c;  or 

c  =  /  (W2  -  Wi)/iW  +  Wi-\-  W2), 

from  which  c  can  be  computed  easily. 

*  Trans.  Am.  Soc.  C.E.,  Vol.  32,  p.  99  (1894). 


Art.  52  27 1 

Fig.  409  represents  in  principle  the  device  used  by  Crandall.  There  were 
two  rollers  under  load  (and  a  third  one  to  preserve  stability  only),  and  three 
plates  as  shown.  The  lower  plate  was  supported  on  the  weighing  table  of  a 
testing  machine;  load  was  applied  on  the  upper  plate;  and  then  the  middle 
plate  was  subjected  to  a  force  P  sufficient  to  start  the  plate.  Thus  the  middle 
plate  was  subjected  to  the  reactions  of  the  two  main  rollers,  inclined  as  shown. 

I  Plate  I 


n 


w 


Plate  \-^^  I  '  '  — 1-^> 


"'^'//////////////A' 


Plate      — I  ■'^'//?^/i(J/////W///////7//tp////////////m/^///////' 


Fig.  409  Fig.  410 

Let  R  =  these  reactions  (nearly  equal),  and  Q  =  their  inclination  to  the  vertical. 
Then,  evidently,  P  =  2  i?  sin  ^  =  2  Rclr^  and  Rco&d  =  W  or  R  =  W  nearly; 
hence 

P  =  2  Wc/r    and    c  =  i  Pr/W. 

Rollers  are  generally  used  for  moving  a  heavy  load  as  shown  in  Fig.  410. 
Let  r  =  radius  of  rollers,  c  =  their  coefficient  of  rolHng  resistance  (assumed 
same  for  top  and  bottom  contacts),  Ri,  R2,  etc.,  =  the  reactions  of  the  rollers, 
6  =  their 'inclinations  to  the  vertical,  W  =  load,  and  P  =  the  pull  required  to 
mo\'e  the  load.  Then  since  6  is  small,  (i?i  +  -^2  +  •  .  .  )  =  W  (nearly) ; 
and  since  sin  d  =  c/r,  P  =  (Ri -{-  R2 -\-     .     .     .     )  c/r.     Hence 

P  =  Wc/r. 

§  2.  Rolling  Wheel.  —  The  general  nature  of  rolling  resistance  in  the 
case  of  a  wheel  is  hke  that  against  a  roller.  A  rolHng  wheel  of  a  vehicle  ex- 
periences axle  friction  as  well  as  rolling  resistance,*  and  few  experiments  have 
been  made  to  determine  them  separately.  For  cast-iron  wheels  20  inches  in 
diameter  on  cast-iron  rails  Weisbach  and  Rittinger,  respectively,  found  for  the 
coefficient  of  rolling  resistance  c  =  0.0183  and  0.0193  inches.f  For  an  iron 
railroad  wheel  39.4  inches  in  diameter,  Pambour  gives  c  =  0.0196  to  0.0216 
inches. 

(i)  Wheel  without  Axle  Friction.  —  We  assume  the  velocity  to  be  constant. 
Of  course  a  force  must  be  applied  to  the  wheel  to  maintain  the  velocity; 
we  assume  it  to  be  applied  to  the  axle  of  the  wheel  as  shown  in  Fig.  411,  and, 
for  simplicity,  that  the  axle  is  frictionless.  Let  D  =  diameter  of  wheel, 
P  =  driving  force,  W  =  weight  of  wheel  and  load  upon  it,  R  =  reaction  of 
roadway,  and  Rh  and  7?„  =  the  horizontal  and  vertical  components  of  R  (see 
Fig.  411b  and  c).     Rh  is  the  "rolling  resistance." 

*  See  Baker's  Roads  and  Pavements  for  full  information  on  total  resistance  to  traction  oi 
vehicles  (due  to  roUinR  resistance  and  axle  friction). 
t  Coxe's  translation  of  Weisbach's  Mechanics. 


272 


Chap,  xii 


Since  there  is  no  angular  acceleration,  the  (resultant)  torque  on  the  wheel 
equals  zero  (see  Art.  51)  and  R  acts  through  the  center  of  the  wheel.     It 


Fig.  411 

follows  that  the  line  of  action  of  R  cuts  the  rim  of  the  wheel  in  front  of  the 
vertical  diameter  of  the  wheel  as  shown,  and  of  course  within  the  arc  of  con- 
tact of  wheel  and  roadway.  The  distance  of  this  point  A  on  the  rim  to  the 
vertical  diameter  is  the  coefficient  of  rolling  resistance,  already  denoted  by  c 
Because  the  torque  of  the  three  forces  (P,  W,  and  R)  is  zero,  and  the  vertical 
distance  between  A  and  0  is  ^  D  nearly,  P  •  |  Z)  =  Wc;  and  since  Rh  =  P, 

R^  =  P=  W  2  c/D.  (2) 

The  work  required  to  overcome  the  rolling  resistance  per  turn  of  the  wheel 

is  equal  to  the  work  done  by  the  driving  force  P  per  turn.     But  this  latter  is 

plainly 

PirD,     or     W  2  TTC.  (3) 

(ii)  Wheel  with  Axle  Friction.  —  Fig.  412  represents  the  wheel,  of  a  horse- 
drawn  vehicle  say,  moving  at  a  constant  speed  toward  the  right.     In  addition 


to  the  foregoing  notation,  let  d  =  diameter  of  axle,  W  =  weight  of  wheel,  Q  = 
resultant  pressure  of  the  axle  on  the  wheel,  Qh  and  Qv  the  horizontal  and  ver- 
tical components  of  Q,  and  /  the  coefficient  of  axle  or  journal  friction  (see 
Art.  45).  R  and  Q  act  somewhat  as  shown  but  their  inclinations  to  the  ver- 
tical are  much  exaggerated.  Fig.  412b  shows  R  resolved  at  A  into  its  hori- 
zontal and  vertical  components,  and  Q  resolved  at  its  point  of  application 
into  its  normal  and  frictional  components  N  and  F;  F  —  fQ.  Fig.  412c  shows 
Q  replaced  by  Qh  and  Qv  at  the  center  of  the  wheel,  and  a  couple  C;  the 
moment  oi  C  is  F  ^  d.     Since  the  speed  is  constant, 

Rh  =  Qh     and     Rh  i  D^  R,c  +  F\d. 

{Continued  on  page  27 g.) 


Art.  53 


273 


53.   Relative  Motion 

§  I.  Motion  Relative  to  a  Point.  —  We  can  specify  position  of  a  point 
only  by  means  of  a  set  of  reference  axes  or  some  other  equivalent  base  de- 
scribed or  implied  in  the  specification.  Thus  when  we  say  that  Chicago  is 
io|  degrees  west  and  3  degrees  north  of  Washington  —  the  cities  regarded 
as  points  —  we  are  really  specifying  the  position  of  the  former  city  with  refer- 
ence to  the  meridian  and  the  parallel  through  Washington.  But  we  say 
briefly  that  the  specification  is  relative  to  Washington.  So  too  when  we  say 
that  a  moving  ship  A  is  40  miles  east  and  50  miles  north  of  another  ship  -B  at  a 
certain  instant,  we  are  specifying  position  of  A  by  means  of  the  meridian  and 
the  parallel  through  B  at  the  instant  in  question;  but  we  say  that  the  specifi- 
cation is  relative  to  B,  the  coordinate  axes  being  understood.  Being  small 
compared  to  the  distances  mentioned  (40  and  50  miles),  the 
ships  were  regarded  as  mere  points.  If,  however,  the  ships  were 
at  close  quarters,  then  to  describe  the  position  of  A  relative  to 
B  we  would  specify  the  position  of  at  least  two  points  in  A 
(bow  and  stern  for  example)  relative  to  axes  fixed  in  B,  as 
indicated  in  Fig.  415,  say.  Even  if  B  were  turning  about,  we 
would  still  use  those  axes  to  specify  subsequent  positions  of  A 
relative  to  B.     For  the  present  we  will  deal  with  position  (and  '  ^^^ 

motion)  of  points  (or  bodies  regarded  as  mere  points)  relative  to  another  base 
point  —  not  body  —  and  it  should  be  understood  that  the  coordinate  axes, 
though  moving  with  the  base  point,  remain  fixed  in  direction. 

Let  the  points  o,  i,  2,  3,  etc.  (Fig.  416),  on  the  lines  aa  and  bb  be  the  positions 


5 

b 
< 

5 

^ 

3 

0 

/ 

ly 
^ 

^ 

4 

a'^ 

/ 

oV 

»». 

J 

5 

a   w 

^ 

z 

3 

4 1 

4    E 


Fig.  416 


Fig.  417 


(relative  to  a  lighthouse),  say,  of  two  ships  A  and  B,  at  12,  i,  2,  .  .  .  o'clock 
of  a  certain  day;  then  the  lines  are  the  paths  (relative  to  the  lighthouse)  of  the 
ships.  From  these  paths  we  have  made  the  following  tabulation  of  coordinates 
of  the  positions  of  A  relative  to  B.  These  coordinates  if  plotted  on  rectangular 
axes,  representing  a  parallel  and  a  meridian,  determine  the  path  of  A  relative 
to  B  (Fig.  417). 

Time  (hours) 

East  (degrees) 

North  (degrees) 


0 

I 

2 

3 

4 

S 

10 

15 

21 

30 

37 

40 

20 

19 

16 

10 

2 

-6 

274 


Chap,  xn 


Taking  points  o;  i,  2,  etc.,  on  the  line  cc  (Fig.  416)  as  the  positions  (relative 
to  the  lighthouse)  of  a  third  ship  C  at  the  hours  mentioned,  we  have  the  follow- 
ing tabulation  of  the  coordinates  of  the  positions  of  A  relative  to  C  from  which 
the  path  of  A  relative  to  C  (Fig.  418)  was  constructed.  Thus  it  is  clear  that  in 
general  the  path  of  a  moving  point  depends  on  the  point  of  reference  or  base 
point. 

Time  (hours) 

East  (degrees) 

North  (degrees) 


0 

I 

2 

•J 

4 

5 

2 

-3 

4 

12 

18 

20 

0 

12 

20 

23 

22 

IS 

For  another  illustration,  imagine  a  table  (Fig.  419),  several  balls  (for  bear- 
ings) on  the  table,  a  drawing  board  resting  on  the  balls,  another  set  of  balls 
on  the  board,  and  a  sheet  of  plate  glass  resting  on  these  latter  balls.     Imagine 


Glass 


Board 


■  Stiff  Wire  fohold 
Pencil  "a" 
A,  B,  a,  and  b 
are  Lead  Pencils. 


Fig.  419 


also  a  pencil  fastened  to  the  glass  so  that  its  point  A  presses  against  the  board, 
and  another  pencil  fastened  to  the  board  so  that  its  point  B  presses  against  the 
glass.  When  either  board  or  glass  or  both  are  rolled  about  on  the  balls,  the 
pencils  (if  suitable  and  properly  mounted)  trace  lines  on  the  board  and  glass. 
If  the  edges  of  the  board  and  glass  (or  coordinates  axes  through  A  and  B)  are 
kept  in  fixed  directions,  then  A  traces  its  path  relative  to  B,  and  B  traces  its 
path  relative  to  ^. 

By  velocity  of  a  point  relative  to  another  point  is  meant  the  rate  at  which 
the  first  point  is  traversing  its  path  relative  to  the  second  point  at  the  instant 
in  question.  We  regard  this  velocity  as  having  direction,  that  of  the  tangent 
to  the  (relative)  path  at  the  point  corresponding  to  the  instant  in  question. 
By  acceleration  of  a  point  relative  to  another  point  is  meant  the  rate  at  which 
the  velocity  of  the  first  point  relative  to  the  second  is  changing  at  the  instant 
in  question. 

Motion  of  Two  Points  Relative  to  Each  Other.  —  The  velocities  and  acceler- 
ations of  two  points  relative  to  each  other  are  equal  and  opposite.  This 
proposition  may  be  illustrated  by  comparing  the  path  of  ship  A,  of  the  fore- 
going illustration,  relative  to  B  (Fig.  417)  with  the  path  of  B  relative  to  A 
(Fig.  420),  constructed  from  the  following  tabulation  of  the  positions  of  B 
relative  to  A  at  the  hours  12,  i,  2,  etc.     Thus  for  the  hour  from  2  to  3  o'clock, 


Time  (hours)  .  .  . 
East  (degrees) ,  . 
North  (degrees) 


0 

I 

2 

3 

4 

—  10 

-15 

—  21 

-30 

-37 

—  20 

-19 

-16 

—  10 

—  2 

5 

-40 

6 


for  example,  the  displacement  of  A  relative  to  B  is  represented  by  vector  2-3 
of  Fig.  417,  and  the  displacement  of  B  relative  to  A  is  represented  by  vector 


Art.  53 


275 


■-E 


Fig.  420 


2-3  of  Fig.  420.  Apparently  these  vectors  are  equal  and  parallel  (also  opposite) ; 
and  it  seems  that  such  displacement  vectors  would  be  equal,  parallel,  and 
opposite  for  any  interval  of  time.  If  this  be  true,  then 
it  follows  that  the  rates  at  which  these  displacements  occur 
(the  relative  velocities)  are  equal  and  opposite  at  each 
instant;  and  if  the  velocities  are  always  equal  and  op- 
posite then  their  rates  of  change  (the  relative  acceler- 
ations) are  also  equal  and  opposite  at  each  instant. 

To  prove  that  displacements  such  as  mentioned  in  the 
preceding  illustration  are  equal  and  opposite,  we  will  use 
the  glass-board  illustration.     Suppose  that  the  pencils  A  and  B  are  attached 
at  the  middle  points  of  the  glass  and  board  respectively,  and  that  at  a  certain 

instant  glass  and  board  are  in  the  positions  shown  at 
(i)  in  Fig.  421,  and  at  a  later  instant  in  positions  shown 
at  (2);  the  table  is  not  shown.  Ai  and  Bi  and  A2  and 
A)  B2  are  the  corresponding  positions  of  the  pencil  points. 
During  this  displacement,  A  will  have  traced  some  such 
line  as  A'A2  and  B  the  line  B'B2.  A^A'  is  equal  and 
parallel  to  B1B2;  hence  A1B1B2A'  is  a  parallelogram, 
and  A'B2  and  AiBi  are  equal  and  parallel.  BiB'  is 
equal  and  parallel  to  ^1^12;  hence  BiA^A^B'  is  a  par- 
allelogram, and  B'A2  and  BiAi  are  equal  and  parallel. 
It  follows  that  A'B2B'A2  is  a  parallelogram,  and  so 
A'A2  and  B'B2  are  equal  and  parallel.  That  is,  the 
displacement  of  A  relative  to  B  (chord  A'Ai)  is  equal 
and  parallel  to  the  displacement  of  B  relative  to  A  (chord  B'Bi).  Obviously 
the  senses  of  the  displacements  are  opposite. 

Motions  of  Two  Points  Relative  to  a  Third  Point.  —  For  convenience  we  re- 
gard the  third  point  as  fixed,  and  call  velocities  and  accelerations  relative  to 
that  point  as  absolute.  To  illustrate  this  case  we  will  modify  the  glass- 
board  apparatus  as  follows:  Imagine  another  pencil  a  rigidly  fastened  to  the 
glass  plate  so  it  presses  against  the  table  as  shown  directly  under  A,  and  B 
extended  downward  so  that  its  lower  end  b  presses  on  the  table.  Then  when 
the  glass  and  board  are  moved  about  without  turning,  a  and  b  draw  the  paths 
of  A  and  B  relative  to  any  (third)  point  as  C  on  the  table;  and  as  already 
stated,  A  and  B  draw  their  paths  relative  to  each  other. 

In  this  case  two  problems  arise:  (a)  Given  the  velocity  (or  acceleration)  of 
a  point  relative  to  a  second  point,  and  the  absolute  velocity  (or  acceleration) 
of  the  second;  required  the  absolute  velocity  (or  acceleration)  of  the  first 
point,  (b)  Given  the  absolute  velocities  (or  accelerations)  of  two  points; 
required  the  velocity  (or  acceleration)  of  either  of  the  two  points  relative  to 
the  other. 

(a)  To  do  this  problem  we  merely  need  to  add  (vectorially) ,  or  compound,  the 
velocity  (or  acceleration)  of  the  first  point  relative  to  the  second  and  the  abso- 


421 


276  Chap,  xii 

lute  velocity  (or  acceleration)  of  the  second;  the  sum  is  the  desired  quantity. 
To  justify  this  solution  we  first  show  that  the  (vector)  sum  of  the  displace- 
ment of  the  first  point  relative  to  the  second  and  the  absolute  displacement 
of  the  second  point  equals  the  absolute  displacement  of  the  first,  all  dis- 
placements being  taken  for  any  interval  of  time.  It  will  follow  that  the  rela- 
tive and  absolute  velocities  (and  accelerations)  are  related  as  above  stated. 
Referring  to  our  glass-board-table  device,  let  A ,  B,  and  C  be  the  three  points 
respectively.  Let  (i)  and  (2),  Fig.  421,  be  the  positions  of  glass  and  board 
at  the  beginning  and  end  of  any  interval,  as  before.  Then  A'Ao  is  the  dis- 
placement of  A  relative  to  B  as  explained;  ^1-62  is  the  absolute  displacement 
of  B;  and  A1A2  is  the  absolute  displacement  of  ^.  As  already  shown,  the 
quadrilaterals  in  the  figure  are  parallelograms;  hence  the  vector  sum  oi  A'A2 
and  B1B2  equals  A1A2. 

(b)  Let  A  and  B  be  the  first  two  points  and  C  the  third,  and  the  velocity 
(or  acceleration)  of  A  relative  to  B  the  desired  quantity.     According  to  (a), 

the  absolute  velocity   (or  acceler- 
ation) of  A  =  the  vector  sum  of 

the  velocity  (or  acceleration)  of  A 

relative    to    B    and    the    absolute 

velocity    (or    acceleration)    of    B. 

Therefore  the  (desired)  velocity  (or 

acceleration)  is  such  a  velocity  (or 

acceleration)  which  when  added 
vectorially  to  the  absolute  velocity  (or  acceleration)  of  5  =  the  absolute 
velocity  (or  acceleration)  of  A.  For  example  let  Va  and  Vb  (Fig.  422)  be  the 
absolute  velocities  (or  accelerations)  of  A  and  B;  then  if  OM  and  ON  be 
drawn  to  represent  Va  and  %  respectively,  NM  will  represent  the  velocity  (or 
acceleration)  of  A  relative  to  B. 

The  problem  can  be  solved  also  on  the  basis  of  the  principle  that  if  we  add 
equal  velocities  (or  accelerations)  to  the  absolute  velocities  (or  accelerations) 
of  the  two  points  we  do  not  change  the  velocities  (or  accelerations)  of  either 
of  the  points  relative  to  the  other.  Thus,  taking  the  preceding  example,  we 
will  add  to  Va  and  Vb  a  velocity  equal  and  opposite  to  Vb  (Fig.  423);  then 
the  new  Vb  =  o  and  the  new  Va  =  NM.  Since  now  B  is  at  rest  relative 
to  C,  the  new  velocity  of  A  relative  to  C  is  also  the  velocity  of  A  relative 
to  B. 

§  2.  Motion  or  a  Point  Relative  to  a  Body.  —  As  explained  in  §  i,  we 
specify  the  positions  of  a  moving  point  relative  to  another  moving  point  by 
means  of  reference  axes  of  fixed  directions  through  the  second  point,  but  its 
positions  relative  to  a  moving  body  by  means  of  reference  axes  fixed  in  the 
body.  See  illustrations  of  the  ships.  Then  the  path  of  a  point  relative  to 
a  body  is  the  fine  through  the  successive  positions  of  the  point  relative  to 
the  body.  Thus,  to  illustrate,  consider  again  the  glass-board-table  appara- 
tus (Fig.  419).    When  both  the  glass  and  board  are  rolled  about  in  any  way, 


/■/ 

Vb 

4 

^^^. 

*-jb. 

/    ho 

1 

-Vb 

N^       / 

•C 

^J>-  / 

^*A 

Fig. 

■  423 

Art.  53 


277 


the  pencil  A  traces  a  line  on  the  board,  and  that  line  is  the  path  of  A  relative 
to  the  board. 

By  velocity  of  a  point  relative  to  a  moving  body  is  meant  the  rate  at  which 
the  point  traverses  its  path  relative  to  the  body  at  the  instant  in  question. 
By  acceleration  of  a  point  relative  to  a  moving  body  is  meant  the  rate  at 
which  the  velocity  of  the  point  relative  to  the  body  is  changing  at  the  instant 
in  question. 

When  a  point  P  is  moving  relative  to  a  moving  body  B  then  the  absolute  velocity 
of  P  equals  the  vector  sum  of  its  relative  velocity  and  the  absolute  velocity  of  that 
point  of  B  with  which  P  coincides  at  the  instant  in  question.  For  simplicity  of 
proof  we  take  the  pencil  A  of  the  glass-board-table  apparatus  as  the  moving 
point  P  and  the  board  as  the  moving  body  B.  Since  P  and  B  have  plane 
motion,  the  proof  is  not  general.     Let  Bdi  (Fig.  424)  be  the  position  of  B  at 


Fig.  424 


a  particular  time  /i,  and  Bdi  the  position  of  5  at  a  later  time  ^2;  also  Pi  and  P2 
respectively,  the  positions  of  P  at  those  times.  Let  M  be  the  point  of  B  with 
which  P  coincides  at  time  ti.  At  time  /i,  M  is  at  Mi  (under  Pi) ;  and  at  time 
/2,  M  is  at  M2.  Then  for  the  interval  t^  —  ti  the  absolute  displacement  of  P 
is  PiP2;  the  relative  displacement  of  P  is  M2P2;  and  the  absolute  displace- 
ment of  M  is  M1M2.  Obviously  P1P2  =  M'2P2  +  MiM^  (vectorially).  Since 
this  relation  holds  for  any  interval,  the  rates  at  which  these  displacements  occur 
(velocities)  are  related  in  the  same  way;  that  is,  the  absolute  velocity  of  P  = 
its  relative  velocity  +  the  velocity  of  M. 

When  a  point  P  is  moving  relative  to  a  moving  body  B  then  the  absolute  accelera- 
tion of  P  equals  the  vector  sum  of  three  accelerations,  namely  —  the  relative  accelera- 
tion of  P,  the  absolute  acceleration  of  that  point  of  B  with  which  P  coincides  at  the 
instant  in  question,  and  a  so-called  complimentary  acceleration.  The  compli- 
mentary acceleration  equals  twice  the  product  of  the  relative  velocity  of  P  and 
the  angular  velocity  of  B  at  the  instant  in  question;  its  direction  is  the  same 


278 


Chap,  xii 


as  that  of  the  Hnear  velocity  of  p  where  Pp  is  a  vector  representing  the  relative 
velocity  of  P  due  to  the  angular  velocity  of  B. 

For  simphcity  again  we  restrict  the  proof  to  plane  motions.  Let  Pipi 
(Fig.  424)  =  the  relative  velocity  of  P  at  the  time  h,  and  If  iWi  ^  the  absolute 
velocity  of  M  at  that  instant.  The  vector  simi  of  these  two  velocities 
equals  the  absolute  velocity  of  P  at  the  time  h.  Making  OA'  and  OB'  to 
represent  these  velocities  respectively,  we  get  the  diagonal  OC  to  represent 
the  absolute  velocity  of  P  at  the  time  h.  Let  N  be  the  point  of  the  board  with 
which  P  coincides  at  the  time  k;  N  is  under  P2  then.  The  velocity  of  N  (at 
time  k)  equals  the  vector  simi  of  the  velocity  of  i/2  and  the  velocity  of  TV 
"about  "  M2.  Now  the  velocity  of  A'  about  I/2  equals  the  product  of  M-2.N 
and  the  angular  velocity  of  the  board  (at  time  h),  or  Ar  X  C02,  where  Ar  =  MoN 
and  C02  =  the  angular  velocity.  The  direction  of  this  velocity  Ar  •  C02  is  perpen- 
dicular to  M2N  as  indicated  (assuming  that  C02  is  counter-clockwise).  OB 
and  bB"  are  equal  and  parallel  to  Mim^  and  Ar'C02  respectively;  hence 
OB"  is  the  velocity  of  N  at  time  h.  Now  let  Po_p-i  (=  OA")  be  the  relative 
velocity  of  P  at  time  h-  Then  the  diagonal  OC"  of  the  parallelogram  on  OA" 
and  OB"  is  the  absolute  velocity  of  P  at  time  to.  Therefore  C'C"  is  the  incre- 
ment in  the  absolute  velocity  of  P  for  the  interval  h  —  h-  It  follows  readily 
from  the  geometry  of  the  figure  that 

C'C"  =  A'A"  +  B'B",  (i) 

vectorial  addition  being  understood  here  and  in  the  following. 

Now  let  Mia  =  P\pi  and  the  angle  between  these  vectors  equal  the  angular 
displacement  Ad  of  the  board  during  the  interval  /2  —  k-  Then  the  increment 
in  the  relative  velocity  of  P  for  that  interval  equals  the  difference  between  the 
vectors  Mia  and  Pipi-  Oa  is  equal  and  parallel  to  Mia;  hence  aA"  is  that 
difference.     Therefore 

A'A"  =  A'a+  Avr  =  2Vr  sin  i  A9  +  Avr, 

where  Vr  means  relative  velocity  of  P  at  time  h.  Since  Ob  is  equal  and  parallel 
to  Mimi  (velocity  of  M  at  time  /o),  B'b  is  the  increment  in  the  velocity  of  M 
during  the  interval  h  —  h',  and  since  bB"  =  Ar'Ui, 

B'B"  =  A%n  +  Ar  •  C02, 

where  Vm  means  velocity  of  M.  Substituting  the  foregoing  values  of  A'A" 
and  B'B"  in  equation  (i),  we  get 

C'C"  =  Azv  +  Avr^  +2Vr  sin  |  A^  -j-  Ar  -  on.  (2) 

Now  let  At  =  h  —  h,  and  ^  approach  ti,  then  we  get 

,.     C'C"      ,.     At..  ,   ,.     Av^n  ,       ,.     Ad  ...     Ar 
hm— 7—  =  lim  -— -  +  hm— -  +  Vr  lim  -—  +  lim  — (02. 
At  At  At  At  At 

The  left-hand  member  is  the  absolute  acceleration  of  P;  the  first  term  of  the 
right-hand  member  is  the  relative  acceleration  of  P;  the  second  term  is  the 
acceleration  of  M.     Lim  {A6/ At)  =  wi,  the  angular  velocity  of  the  board  at 


Art.  53  279 

time/i;  hence  the  third  term  =  ZJrCOi.  Lim  (Ar/A/)co2  =  lim  (Ar/At)  X  limw2 
=  Vr^i.  Hence  the  third  and  fourth  terms  are  equal  in  magnitude,  and  if  their 
directions  —  they  are  vectors  —  are  parallel,  then  their  smn  =  2  VrWi.  The 
direction  of  the  third  term  is  the  limiting  direction  of  A' a,  perpendicular  to 
OA'  or  Vr  obviously.  The  direction  of  the  fourth  term  is  the  limiting  direction 
of  bB"  or  Nc.  Now  Nc  is  always  (as  /2  approaches  h)  perpendicular  to  M2P2; 
and  since  1/2^2  is  the  relative  displacement  of  P,  the  Umiting  direction  of 
M2P2  is  Pipi  (or  Vr).  Hence  the  limiting  direction  of  Nc  is  perpendicular  to  Vr. 
Thus  the  sum  of  the  last  two  terms  =  2  Vrcoi,  and  it  has  the  direction  mentioned, 
perpendicular  to  tv-  And  this  sum  is  the  so-called  complimentary  acceleration; 
it  is  called  also  acceleration  of  Coriolis  after  him  who  first  discovered  the  rela- 
tion between  the  accelerations  under  discussion. 

{Continued  from  page  272!) 
From  the  last  equation  on  that  page   and  Rv  =  Qv  ^-  W,  it  follows  that 

Rh  =  Qh={Q.  +  W')^  +  F^-  (4) 

In  most  cases  of  vehicles,  Qv  is  nearly  equal  to  Q,  and  W  is  neghgible  com- 
pared to  Qv ;  therefore  we  may  write  as  a  close  approximation 

The  work  required  to  overcome  rolling  resistance  and  axle  friction  is  fur- 
nished by  Qh.     Per  turn  of  the  wheel,  that  work  is 

QtjvD  =  ((?„  +  W)  2  TC  +  Qhd  =  TT  (2  c  +  /J)  (?.  (5) 


CHAPTER   XIII 

THREE  DIMENSIONAL    (SOLID)    MOTION 

54.  Body  with  a  Fixed  Point;  Kinematics 

§  I.   Spherical  Motion  means  motion  of  a  rigid  body  with  only  one 

point  of  the  body  fixed.     Each  point  of  the  body,  excepting  the  fixed  one, 

moves  on  the  surface  of  a  sphere,  whence  the  name  spherical  motion. 

Any  spherical  displacement  of  a  body  can  be  accomplished  by  means  of  a 

rotation  about  some  line  of  the  body  passing  through  the  fixed  point,  and 

fixed  in  space.     Proof:  —  Evidently,  we  mav  describe  any  position  of  the 

body  by  describing  the  positions  of  two  of  its  points,  not  in  line  with  the  fixed 

point.     Let  A  and  B  denote  two  such  points,  equally  distant  from  the  fLxed 

point  0;   then  during  any  motion  of  the  body,  A  and  B  move  on  the  surface 

of  the  same  sphere.     Let  OAiBi  be  one  position  of 

the  body,  and  OA2B2  another.     Then  we  are  to 

prove  that  the  points  A  and  B  could  be  brought 

from  AiBi  to  .42^2  by  means  of  a  single  rotation 

about  some  fixed  line  through  0.     Let  the  lines 

.4 1^1  (Fig.  425)  and  A2B2  be  arcs  of  great  circles 

of  the   sphere   mentioned;    these   arcs   are   equal 

since  A  and  B  are  points  of  a  rigid  body.     The 

fines  A1A2  and  B1B2  are  arcs  of  great  circles;  M 

_  and  N  bisect  these  arcs;  MR  and  NR  are  great 

Fig  a2^ 

circles  perpendicular  to  ^1-42  and  B1B2  respectively. 

In  general  two  such  great  circles  do  not  coincide  but  intersect  at  two  points,  R 
and  S.  The  diameter  ROS  is  the  axis,  rotation  about  which  would  produce 
the  given  displacement,  proven  presently.  Let  AiR,  A2R,  BiR,  and  B2R  be 
arcs  of  great  circles.  Since  A1A2R  and  B1B2R  are  isosceles  triangles,  AiR  = 
A2R  and  BiR  =  BoR;  and,  as  already  stated,  AiBi  =  .42^2.  Hence  the  trian- 
gles RAiBi  and  RA2B2  are  equal,  and  the  angle  AiRBi  =  A2RB2.     Finally, 

A1RA2  =  A1RB2  —  A2RB2  =  A1RB2  -  AiRBi  =  B1RB2. 

Hence  a  rotation  of  the  great  circles  A^R  and  BiR  about  RS  of  an  amoimt 
equal  to  the  angle  A1RA2  would  displace  A  from  ^1  to  ^2  and  B  from  Bi  to  B2. 
Imagine  any  actual  continuous  spherical  motion  of  a  body,  in  which  the 
two  points  A  and  B  of  the  body  are  displaced  from  ^1  to  yl2  and  Bi  to  B2  re- 
spectively. Let  A',  A",  etc.,  be  several  intermediate  positions  of  A,  and  let 
B',  B",  etc.,  be  corresponding  intermediate  positions  of  B.  As  already  shown, 
the  displacements  of  AB  from  AiB^  to  A'B',  from  A'B'  to  A"B",  from  A"B" 

280 


Art.  54  281 

to  A"'B"',  etc.,  might  be  accomplished  by  single  rotations  about  definite 
fixed  lines  R'OS',  R'VS",  R"'OS"',  etc.  If  a  large  number  of  intermediate 
positions  A'B',  A"B",  etc.,  be  assumed,  and  if  the  successive  rotations  be 
accomplished  in  times  equal  to  the  times  required  for  the  actual  displace- 
ments in  the  continuous  motion,  then  the  succession  of  rotations  would  closely 
resemble  the  actual  continuous  motion.  The  more  numerous  the  interme- 
diate positions,  and  the  more  numerous  the  succession  of  single  rotations, 
the  more  closely  would  the  succession  resemble  the  actual  motion.  "In  the 
limit,"  the  succession  would  reproduce  the  actual  motion;  hence  we  may 
regard  any  spherical  motion  of  a  body  as  consisting  of  a  continuous  rotation 
about  a  line  through  the  fixed  point,  the  fine  continually  shifting  about  in  the 
body  and  in  space.  The  Kne  about  which  the  body  is  rotating  at  any  instant 
is  the  instantaneous  axis  (of  rotation)  at  that  instant. 

At  any  particular  instant  of  a  spherical  motion,  the  body  is  rotating  about 
the  instantaneous  axis  at  a  definite  rate;  this  rate  is  called  the  angular  velocity 
of  the  body  at  that  instant.  We  will,  generally,  denote  magnitude  of  angular 
velocity  by  co.  In  a  rotation  about  a  fixed  axis,  the  (linear)  velocity  of  any 
point  of  the  body  equals  the  product  of  the  angular  velocity  and  the  perpen- 
dicular distance  (or  radius)  from  the  point  to  the  axis;  and  the  direction  of 
the  Unear  velocity  is  perpendicular  to  the  plane  of  the  radius  and  the  axis.  So 
too  in  a  spherical  motion,  the  linear  velocity  of  any  point  of  the  body  at  any 
instant  equals  the  product  of  the  angular  velocity  at  that  instant  and  the 
radius  (perpendicular  from  the  point  to  the  instantaneous  axis  for  that  in- 
stant); the  direction  of  that  velocity  is  perpendicular  to  the  plane  of  the 
radius  and  the  axis. 

Any  angular  velocity  oj  may  be  represented  by  means  of  a  vector  laid  off 
on  the  corresponding  instantaneous  axis;  the  length  of  the  vector  is  made 
equal  to  co  according  to  some  convenient  scale,  and  the  sense  of  the  vector 
indicates  the  direction  of  the  rotation  according  to  some  convention.  We 
will  always  associate  direction  with  angular  velocity  in  the  way  just  described; 
that  is,  we  regard  angular  velocity  as  a  vector  quantity.  In  a  spherical  mo- 
tion, angular  velocity  changes  in  direction  continuously;  it  may  or  may  not 
change  in  amount  too.  In  any  case,  the  rate  at  which  the  (vector)  angular 
velocity  is  changing  at  any  instant  is  called  the  angular  acceleration  at  that 
instant.  (See  page  148  for  note  on  rate  of  change  of  a  vector  quantity.)  This 
rate  or  acceleration  has  a  definite  amount  and  direction  at  each  instant,  and 
hence  is  a  vector  quantity  too.  We  will  use  a  to  denote  the  magnitude  of  an 
angular  acceleration. 

§  2.  Composition  and  Resolution  op  Angular  Velocities.  —  Imagine 
a  body  P  to  be  rotating  about  a  line  li  fixed  in  a  body  A  ;  and  that  A  is  rotating 
about  a  line  h>,  intersecting  h  and  fixed  in  a  body  B  (Fig.  426).  For  conve- 
nience we  call  the  motion  of  P  relative  to  B  its  absolute  motion,  and  we  regard 
this  absolute  motion  as  a  resultant  motion  consisting  of  the  (component) 
rotations  about  h  and  h-    We  will  show  presently  that  the  absolute  motion 


282 


Chap,  xiii 


of  P  is  spherical,  and  that  the  angular  velocity  of  that  motion  equals  the 
vector  sum  of  the  angular  velocity  of  P  relative  to  A  and  that  of  A  relative 
toB. 

That  the  absolute  motion  is  spherical  will  be  conceded  as  almost  self-evident; 
for  the  point  0  (of  P)  does  not  move  at  all,  being  a  point  of  B,  and  O  appears 
to  be  the  only  point  of  P  which  is  fixed.     But  more  on  this  matter  later. 

Let  Oa  and  Oh  (Fig.  427)  be  the  two  lines  h  and  h  at  the  instant  in  question, 
and  let  the  angular  velocities  (of  rotation)  about  those  lines  be  coi  and  C02  re- 
spectively, and  in  the  directions  indicated.  Let  C,  not  shown,  be  any  point 
of  the  body  P  in  the  plane  of  the  lines  h  and  h  (or  paper).     If  C  is  taken  above 


Fig.  426 


Oa,  then  the  rotation  coi  alone  brings  C  up  out  of  the  paper;  if  below,  then 
cui  depresses  C.  If  C  is  above  Oh,  then  C02  alone  brings  C  up  out  of  the  paper; 
if  below,  002  depresses  C.  Hence  if  C  is  in  either  acute  angle  between  h  and  h, 
the  two  rotations  give  C  displacements  in  opposite  directions.  Let  Co  be 
such  a  point  C,  and  so  chosen  too  that  the  two  displacements  of  Co  in  an  ele- 
ment of  time  dt  would  be  equal.  If  ri  and  r^  =  the  distances  of  Co  from  h 
and  h  respectively,  these  displacements  =  ricoi  dt  and  r^2  dt.  Hence,  /iwi  = 
^2^2,  or  OCo  sin  a-on  =  OCo  sin  /3  •  wo;  and 


sin  a-  coi  =  sin  j8  •  ooo,     or  sin  a/ sin  (3  =  0)2/0)1. 


(i) 


Let  D  be  any  other  point  on  OCo;  then  its  displacements  due  to  oji  and  0)2  in 
the  time  dt  are  respectively  (OD  sin  a)  wi  dt  and  {OD  sin  /3)o)2  dt.  But  these 
are  equal,  since  sin  a  •  oji  =  sin/3  •  0)2;  hence  all  points  on  OCo  have  zero  velocity 
at  the  instant  in  question.  Evidently,  there  are  no  other  points  in  the  body 
P  whose  velocity  is  zero  at  the  instant;  hence  the  state  of  motion  of  P  is  a 
rotation  about  OCo,  a  Hne  fixed  by  equation  (i). 

Let  0)  =  the  angular  velocity  of  the  rotation  of  P  (about  OCo);  Q  (Fig.  427) 
be  any  point  of  P  in  the  plane  of  the  paper;  q,  gi,  and  g2  =  distances  of  Q  from 
OCo,  h,  and  k  respectively.  Then  the  displacements  of  Q  due  to  o)i  and  0)2 
are  respectively  gio:i  dt  and  ^2^2  dt.  These  displacements  for  Q  as  chosen  are 
in  the  same  direction;  hence  the  total  or  resultant  displacement  =  (910)1  -f- 
92W2)  dt,  and  the  linear  velocity  of  Q  (displacement  per  unit  time)  =  ^lOJi  + 
5'2Wp,.    Now  the  angular  velocity  of  the  body  P  equals  the  linear  velocity  of 


Art.  54 


283 


its  point  Q  divided  by  the  distance  of  Q  from  the  (instantaneous)  axis  of  rota- 
tion OCo,  or 

CO  =  (^lOJi  +  q-iOii)  -^  q. 

It  follows  from  the  trigonometry  of  the  figure  that 

gi  =  q  cos  a  +  (OR)  sin  a,     and  q-z  =  q  cos  /3  —  (OR)  sin  /3. 

Substituting  these  values  of  qi  and  q^  in  the  expression  for  co  and  noting  equa- 
tion (i),  we  arrive  at 

CO  —  wi  cos  a  +  C02  cos /3.  (2) 

Equations  (i)  and  (2)  respectively  enable  us  to  determine  the  axis  (OCo)  of 
the  resultant  of  two  angular  velocities  coi  and  coo  and  the  amount  of  the  result- 
ant angular  velocity  co. 

By  means  of  equations  (i)  and  (2)  we  can  show  that  co  is  the  vector-sum  of 
coi  and  C02.  Let  OM  and  ON  (Fig.  428),  on  h  and  k  (Figs.  426  and  427),  repre- 
sent 0)1  and  C02,  and  let  OMNR  be  a  parallelogram.     Then 

coi  sin  MOR  =  C02  sin  NOR. 

Comparing  this  with  equation  (i),  we  see  that  MOR  =  a  and  NOR  =  /3; 
hence  the  parallelogram  construction  gives  a  diagonal  which  coincides  with 


J^-z 


Fig.  428 


Fig.  429 


the  instantaneous  axis  of  the  absolute  motion.  It  will  be  readily  seen  from 
the  parallelogram  that 

OR  =  coi  cos  a  -\-  C02  cos  (3 ; 

hence  the  length  of  the  diagonal  gives  the  magnitude  of  the  angular  velocity 
(see  equation  2). 

Reverting  now  to  the  proposition  that  the  absolute  motion  of  P  is  spher- 
ical, we  note  that  the  axis  of  instantaneous  rotation  OCo  (Fig.  427)  is  always 
in  the  plane  of  /i  and  h  and  hence  is  not  fixed.  Therefore  there  is  only  one 
fixed  point  of  P,  the  intersection  of  /i  and  h. 

Obviously,  the  foregoing  analysis  could  be  extended  to  a  case  of  simulta- 
neous rotations  about  three  or  more  concurrent  axes  /i,  I2,  h,  etc.  Hence,  in 
any  case,  the  resultant  motion  is  spherical,  and  the  resultant  angular  velocity 
is  given  by  the  vector-sum  of  the  component  angular  velocities.  Conversely, 
the  angular  velocity  of  any  spherical  motion  can  be  resolved  into  any  number 


284  Chap,  xiii 

of  concurrent  components,  and  the  vector-sum  of  the  components  is  equal  to 
the  given  velocity. 

§  3.  Velocity  of  Any  Point  of  the  Moving  Body. — Let  P  (Fig.  429) 
be  any  point  of  a  moving  body  (not  shown),  fixed  at  0;  x,  y,  and  z  the  (chang- 
ing) coordinates  of  P  with  reference  to  fixed  axes  OX,  OY  and  OZ;  co^,  co^,  and 
o3z  =  the  components  of  the  angular  velocity  of  the  body  with  respect  to  those 
axes;  v  =  the  linear  velocity  of  P;  and  Vx,  Vy,  and  v^  =  the  components  of  v 
along  those  axes.     Then  as  will  be  proved  presently 

I'x  =  ZOOy  —  ycOz,       Vy  =  X(j3z  —  SWx,       Vz  =  JCOx  "  XOiy.  (3) 

If  the  body  were  rotating  about  the  x  axis  only,  then  P  would  be  describing  a 
circle  about  X,  and  the  velocity  of  P  would  be  XP  X  co^.  This  velocity  has 
no  X  component,  and  it  is  plain  from  the  figure  that  the  y  and  z  components 
of  that  velocity  respectively  are  —  zwx  and  jco^.  These  component  velocities 
of  P  due  to  angular  velocity  cox  are  scheduled  below;  also  the  component 
velocities  due  to  angular  velocities  cOy  and  w^.  It  is  plain  from  the  schedule 
that  the  total  component  velocities  due  to  the  three  angular  velocities  are  as 
given  by  equations  (3). 

Rotation  about  OX  produces 
OY  produces 
OZ  produces 

55.   Body  with  a  Fixed  Point;  Kinetics 

§  I.  Angular  Momentum.  —  We  will  now  explain  what  is  meant  by  an- 
gular momentum  of  a  body  about  or  with  respect  to  its  fixed  point.  (For 
meaning  of  angular  momentum  about  a  line  see  Art.  48.)  By  angular  mo- 
mentum of  a  particle  about  a  point  is  meant  the  moment  about  that  point  of 
its  momentum.  Thus  if  m  and  v  are  the  mass  and  velocity  of  a  particle  and 
p  is  the  perpendicular  distance  from  the  point  to  the  line  drawn  through  the 
particle  and  in  the  direction  of  its  motion  at  the  instant  in  question,  then  the 
angular  momentum,  about  the  point,  of  the  particle  at  the  instant  is  mvp. 
(See  Fig.  430.)  This  angular  momentum  is  represented  by  a  vector  which 
we  call  r,  through  the  point  0,  perpendicular  to  the  plane  of  O  and  the  mo- 
mentum vector  mv;  the  length  of  r  is  made  equal  (according  to  some  conven- 
ient scale)  to  mvp,  and  the  arrow  on  the  vector  is  fixed  in  accordance  with  the 
usual  rule.  By  angular  momentum  of  a  body  about  its  fixed  point  is  meant 
the  resultant  of  the  angular  momentums  about  that  point  of  all  its  particles. 
Thus  the  angular  momentum,  about  the  fixed  point  of  the  body,  is  represented 
by  a  definite  vector  R,  namely,  the  resultant  of  the  vectors  which  represent 
the  angular  momentums,  about  the  point,  of  the  particles  of  the  body. 

If  the  vector  representing  the  angular  momentum  of  a  body  about  the 
fixed  point  be  resolved  into  a  (rectangular)  component  along  any  line  through 
the  point,  such  component  vector  will  represent  the  angular  momentum  of 


Vx  =  0, 

Vy     =      —    ZOJx, 

and 

Vz  =  ycox- 

Vx    =    ZiOy, 

Vy  =  0, 

and 

Vz    =      —    XClOy. 

Vx  =  -  yo:z, 

Vy    =    XWz, 

and 

Vz  =  0. 

Art.  55  285 

the  body  about  the  line.  We  prove  this  proposition  for  a  particle,  and  then 
extend  the  proof  to  a  collection  of  particles,  —  that  is  a  body.  Let  P  (Fig. 
430)  be  one  of  the  particles  of  the  body,  not  shown,  0  the  fixed  point,  OX  the 
line,  r  the  vector  representing  the  angular  momen- 
tum of  P  about  0,  r^  the  component  of  r  along  OX, 
p  the  perpendicular  from  O  to  the  vector  mv,  or  \^^^^ 
PQ,  and  a,  /3,  and  7  the  direction  angles  of  v.    Then  ^\ 

Vi  =  mvp  cos  XON  =  mvp  (cos  y  -  y  —  cos  13- z)  ^  p  ^ 

=  m  (vcosy  '  y  —  vcos^  '  z)  =  mvzy  —  mvyZ.  ^^  y_..--  " 

But  this  last  expression  is  the  value  of  the  angu-  y\q,.  430 

lar  momentum  of  P  about  the  line  OX  (Art.  48). 

Al^o  Sr^  =  2w  {v^y  —  Vyz)     or    R^  =  h^, 

where  R^  denotes  the  component  of  R  along  OX  and  /?x  the  angular  momentum 
of  the  body  about  OX;  thus  the  proposition  has  been  proved. 

Obviously  the  greatest  value  of  Rx  is  R,  and  hence  Jh  is  greatest  when  OX 
coincides  with  R.  That  is  to  say,  with  respect  to  lines  through  the  fixed 
point  of  a  body,  the  angular  momentum  is  greatest  for  the  line  coinciding 
with  the  vector  which  represents  the  angular  momentum  of  the  body  about 
the  point;  this  greatest  angular  momentum,  about  a  line,  equals  the  angular 
momentum  about  the  point.  For  brevity,  we  will  call  these  (equal)  angular 
momentums  the  or  resultant  angular  momentum  of  the  body;  we  will  denote 
it  by  h.  Finally  h^,  hy,  and  hz  will  be  our  symbols,  not  only  for  the  angular 
momentums  of  the  body  about  x,  y,  and  z  axes  through  the  fixed  point  but 
for  the  components  of  h  along  those  lines. 

Referring  to  the  foregoing  or  Art.  48,  it  will  be  seen  that 

hx  =  ^m{vzy-VyZ),    hy  =  '^m{vxZ  -  v^x),    h=^m{vyX  -  v^y).         (i) 

These  expressions  for  the  component  angular  momentums  can  be  transformed 
into  the  following  (involving  angular  and  not  linear  velocities) : 

hx  =    +  Ix^x  —  JzWy  —  Jy(Jiz, 

hy  =    —  J  zWx  +  ly<J^y  —  J  x^y,  \V 

hz   —    —  J^jix  ~  Jxi^y  +  ^zW«. 

/;,,  /„,  and  Iz  =  the  moments  of  inertia  of  the  body  about  the  x,  y,  and  2  axes 
respectively;  or  symbolically, 

/,  =  Sm(/  +  32),     [y  =  ^m(z'-\-x'),     Tz  =  ^m{x'-{-y^). 

J^,  Jy,  and  Jz  respectively  =  the  products  of  inertia  of  the  body  with  respect 
to  the  two  coordinate  planes  intersecting  in  the  x,  y,  and  z  axes  (Art.  57);  or 
symbolically, 

Jx  =  '^myz,    Jy  =  "Zmzx,    Jz  =  Swxy. 

Symbols  oox,  coy,  and  coz  denote  the  axial  components  of  the  angular  velocity 
of  the  body.     Equations  (2)  may  be  deduced  from  equations  (i)  by  substi- 


286  Chap,  xiii 

tuting  for  v^,  Vy,  and  Vy  their  values  from  equations  (3)  of  Art.  54,  and  then 
simplifying.  If  the  coordinate  axes  x,  y,  and  s  are  principal  axes  of  the  body 
at  the  fixed  point  (Art.  57),  then  /x,  Jy,  and  Jz  =  o;  and 

It  is  worth  noting  that  the  axis  of  rotation  of  a  body  with  a  fixed  point 

(Art.  54),  and  the  vector  of  the  angular  momentum 
about  the  point  are  in  general  not  coincident.     Thus 
let  OA,  OB,  and  OC  be  three  fixed  axes   coincid- 
ing with  the  principal  axes  at  O  of  a  body  (not 
shown)  which  is  fixed  at  O.    If  OA,  OB,   and  OC 
-}^  represent    the  components   (co^,  Wy,  and  coj),   along 
the   principal    axes,   of  the  angular   velocity  co   of 
the  body,  then  OD  represents  w  and  OD  is  the  axis 
of  the  rotation.     If  Oa,  Ob,  and  Oc  represent  hx,  hy, 
and   hz  (TxCJx,   ly^yj  ^^^  J^z^z),  then  Od  represents  h.       And  obviously  OD 
and  Od  coincide  only  if  Jh  :  hy  :  hz  ::  co^  :  ojj^  :  ojz;  that  is  if  /^  =  /"„  =  Iz. 

§  2.  Rate  of  Change  or  Angular  Momentum.  —  If  the  x,  y,  and  s  axes 
are  fixed,  then  expressions  for  the  rates  of  change  of  hx,  hy,  and  hz  may  be 
obtained  by  (ordinary)  differentiation  of  the  right-hand  members  of  equa- 
tions (2).  The  results  arrived  at  will  not  be  simple,  since  in  general  all  /'s, 
/'s,  and  co's  vary  with  the  time.  Simpler  results  for  these  rates  will  now  be 
stated  and  deduced  by  means  of  "moving  axes." 

Let  0  (Fig.  432)  be  the  fixed  point  of  the  moving  body  (not  shown);  OA, 
OB,  and  OC  a  set  of  axes  moving  with  the  body  or  otherwise;  d  the  (vary- 
ing) angular  velocity  of  the  frame  OABC,  and  di,  62,  and  83  the  components 
of  6  along  OA ,  OB,  and  OC  respectively.  If  these  moving  axes  are  fixed  in  the 
body,  then  9  =  co,  di  =  o)x,  62  =  o}y,  and  63  =  ojz.  Let  OX,  OY,  and  OZ  be  a 
set  of  fixed  axes  with  which  the  moving  axes  coincide  as  shown  at  the  instant 
in  question.     Then  at  that  instant,  hi  =  h^,  hp.  =  //.,,,  and  hz  =  hz. 

In  general  all  /z's  vary;  the  corresponding  ones  {]i\  and  hj^,  hi  and  hy,  hz  and 
//z)  do  not  vary  at  equal  rates,  even  at  the  instant  when  the  sets  of  axes  coin- 
cide. As  will  be  shown  presently,  these  rates  at  the  instant  of  coincidence 
are 

-  h^z  +  hdi, 

-  hdi  +  hds, 

dhz      dh       1  Q    X    1  a 

where  the  derivatives,  the  /2's  and  the  0's  all  pertain  to  the  instant  in  question. 

Even  when  the  moving  axes  are  not  coincident  with  the  fixed  set,  hj^  equals 

the  sum  of  the  components  along  OA'  of  hi,  ho,  and  hs.     And  in  general,  changes 

in  amount  or  direction  of  hi,  h^,  and  h^  affect  hx  and  also  hy  and  hz.     Suppose 


dhx 
dt 

_  dhi 
dt 

dhy 
dt 

dh 
dt 

Art.  55 


287 


first  that  the  body,  but  not  the  frame  OABC,  is  rotating;  and  that  the  frame 
coincides  with  OXYZ.  Change  in  Ih  and  I13  would  not  affect  h^;  but  change 
in  hi  would.  The  rate  of  change  in  h^  would  be  dhy/dt.  Now  consider  the 
effect  on  hx  of  any  rotation  of  the  frame  OABC  about  0.  The  component 
velocity  di  turns  Ih  and  Jh  in  the  plane  YOZ,  and  hence  does  not  affect  K. 
The  component  62  turns  hi  and  hs  in  the  plane  ZOX;  the  turning  of  hi  does 


YCbB 


Fig.  432 

not  affect  ^^  but  turning  of  hs  contributes  ^3^2  to  the  rate  of  h^  at  the  instant 
when  the  axes  coincide.  The  component  d^  turns  hi  and  //2  in  the  plane  XOY; 
the  turning  of  h  does  not  affect  h^  but  turning  of  ^  contributes  —  h^Os  to  the 
rate  of  hx  at  the  instant  when  the  axes  coincide.*  Finally,  the  total  rate  of  hx 
is  (dhi/di)  +  /72^3  +  h02  which  was  to  be  shown.  In  a  similar  way,  the  stated 
values  of  the  other  two  rates  could  be  arrived  at. 

Since  the  torque  of  the  external  forces  acting  on  a  body  about  any  fixed 
line  equals  the  rate  at  which  the  angular  momentum  about  that  line  is  chang- 
ing (Art.  48), 

dlh 
dt 


Tx- 


h^dz  +  /?3^2  =  Ti, 


dho 


Ty=Yt-  ^''^'  +  ^1^3  =   ^2, 


T^  =  ^-  hA  +  //29i  =  ^3, 
dt 


(3) 


where  Tx,  Ty,  and  T^  denote  the  torques  about  the  x,  y,  and  2  axes  respec- 
tively, and  Ti,  T2,  and  ^3  those  about  the  coincident  moving  axes.  Torques, 
rates,  etc.,  pertain  to  the  same  instant,  of  course. 

*  Let  OP  (Fig.  433)  be  a  vector  R  of  constant  magnitude  rotating  about  O,  in  the  plane 
of  the  paper,  with  angular  velocity  co  (=  d<j>/dO.  Also  let  R'  =  the  component  of  R  along 
the  fixed  line  OU;    then  R'  =  R  cos  <>.     The  rate  at  which  R'  changes  is 

dR'/dt  =  -  i?  sin  (/)  •  d<t>/dl  =  -  7?  sin  <^  •  «. 

Now  when  R  coincides  with  OU,  </>  =  o,  and  the  rate  of  change  of  R'  =  o.  When  R  is  per- 
pendicular to  OU,  <t>  =  ^TT,  and  the  rate  of  change  of  /?'  =  -  i?w.  If  w  is  positive  (rotation 
counter-clockwise),  then  -  Ru  is  negative;  and  this  means  that  R'  is  decreasing  (obvious 
from  the  figure).  If  w  is  negative  (rotation  clockwise),  then  —  Rw  is  positive;  and  this 
means  that  R'  is  increasing  (obvious  from  the  figure). 


288 


Chap,  xm 


§  3.  Kinetic  Energy.  —  As  in  §  3  of  Art.  54,  let  co^,  o)y,  and  co^  =  the  axial 
components  of  the  angular  velocity  of  the  moving  body  at  any  particular 
instant;  x,  y,  and  z  =  the  coordinates  of  some  particle  P  of  the  body  then; 
and  Vx,  Vy,  and  Vz  =  the  axial  components  of  the  velocity  of  P.  If  m  =  the 
mass  of  the  particle,  then  its  kinetic  energy  at  the  instant  in  question  is 

Now  if  we  substitute  for  Vx,  Vy  and  Vz  their  values  from  equations  (3)  of  Art. 
54,  we  arrive  at  a  new  expression  for  the  kinetic  energy  of  P;  and  if  we  sum 
up  such  expressions  for  all  the  particles  of  the  body,  we  find  that  the  kinetic 
energy  of  the  body  is 


x^x 


J  Z^xWy, 


+  2  I^J^v'  +  \  ^^?  ~  J^yOiz  —  JyOiz(Jix 

where  Ix,  ly,  and  I^  are  the  moments  of  inertia  of  the  moving  body  about  the 
X,  y,  and  2  axes  respectively,  and  Jx,  Jy,  and  J ^  are  the  products  of  inertia  of 
the  body  with  respect  to  the  pairs  of  coordinate  planes  intersecting  in  the 
X,  y,  and  z  axes  respectively  (Art.  57)  all  at  the  instant  in  question.  That  is 
Jx  =  2wyz,     Jy  —  Zmzx,    J^  =  Swxy. 

The  products  of  inertia  may  be  zero ;  then  the  kinetic  energy  equals 


56.   Gyrostat 

§  I.  Steady  Oblique  Precession.  —  Let  OC  (Fig.  434)  be  the  spin-axis 
of  a  gyrostat,  OZ  a  fixed  axis  about  which  OC  is  rotating  or  precessing  at  a 
steady  rate,  d  =  the  constant  angle  ZOC,  w  =  the  angular  velocity  of  spin, 

and  fi  =  the  angular  velocity  of  precession.  Let 
OX  and  OY  be  fixed  axes,  perpendicular  to  each 
other  and  to  OZ;  OA  an  axis  on  the  plane  of  ZOC 
and  perpendicular  to  OC;  and  OB  perpendicular 
to  OA  and  OC.  OA  and  OB  are  moving  axes, 
rotating  with  OC  about  OZ. 

The  motion  of  the  gyrostat  consists  of  the  com- 
ponent rotations  co  and  ^  about  OC  and  OZ  re- 
spectively.    The  resultant  of  those  components  is 
a  rotation  about  the  diagonal  of  the  parallelogram 
on  the  vectors  Oc  and  Oz  representing  w  and  Q, 
(Art.  54),  and  the  angular  velocity  of  that  resultant  rotation  is  represented 
by  that  diagonal.     Hence,  the  components  of  the  angular  velocity  of  the 
gyrostat  along  OA,  OB,  and  OC  are  respectively 

—  12  sin  Q,     o,     and    00  -f  12  cos  6  =  n, 

n  being  an  abbreviation  for  co  +  fi  cos  d. 

It  may  be  well  to  note  the  distinction  between  the  velocity  of  spin  co  and  n. 
The  spin  velocity  is  the  angular  velocity  of  the  gyrostat  relative  to  the  moving 


Fig.  434 


Art.  $6  289 

frame  OABC;  it  is  the  product  of  2  tt  and  the  number  of  times  per  unit  time 
which  a  point  on  the  gyrostat  pierces  the  plane  ZOC.  The  angular  velocity  n 
is  the  component  of  the  absolute  angular  velocity  of  the  gyrostat  along  the 
fixed  line  with  which  OC  happens  to  coincide  at  the  instant  in  question.  Since 
OA,  OB,  and  OC  are  principal  axes  of  the  gyrostat  (Art.  57),  the  angular  mo- 
mentums  of  the  gyrostat  about  these  axes  are  respectively 

—  ^  12  sin  6,     o,     and    C  (co  +  12  cos  6)  =  Cn, 

where  A  and  C  denote  the  moments  of  inertia  of  the  gyrostat  about  OA  and 
OC  respectively. 

Since  the  entire  frame  of  axes  OABC  is  rotating  about  OZ  with  angular 
velocity  12,  the  components  of  that  velocity  along  OA,  OB,  and  OC  are  re- 
spectively, 

—  12  sin  6,     o,     and     12  cos  9. 

Substituting  now  in  equation  (3)  of  Art.  55,  we  get  as  the  required  values  of 
the  torques  of  the  external  forces  about  the  axes  OA,  OB,  and  OC  respectively 

Ti  =  o, 

Tz  =  Cn  12  sin  0  -  /1 12-  sin  d  cos  6, 
and  T3  =  o. 

Therefore  for  steady  spin  and  precession,  there  must  be  no  torque  about  any 

line  in  the  plane  of  the  spin  and  precession  axes  {Ti  and  Tz  =  o)  but  a  torque 

equal  to 

Cn  12  sin  0  -  ^  12-  sin  6  cos  d 

about  a  line  perpendicular  to  those  axes  and  through  the  fixed  point. 

Let  us  now  consider  whether  a  gyrostat  may  precess  steadily  under  the  in- 
fluence of  gravity  and  the  pivot  reaction  only.  Let  W  =  the  weight  of  the 
gyrostat,  and  h  =  the  distance  of  its  center  of  gravity  from  the  pivot;  then  the 
torques  of  gravity  about  OA,  OB,  and  OC  are  respectively  o,  Wh  sin  6,  and  o. 
We  assume  that  the  pivot  is  so  well  made  that  the  torques  of  the  reaction 
about  the  lines  mentioned  equal  zero  practically.  Hence  the  gyrostat  is  not 
subjected  to  any  torque  about  OA  and  OC  but  to  a  torque  Wh  sin  d  about  OB. 
If  now  the  quantities  W,  h,  d,  etc.,  be  given  such  values  that 

Wh  sin 6  =  Cnn  sind  -  A  12=  sin d cos 9, 

then  all  the  conditions  for  steady  precession  will  be  satisfied.  Evidently 
such  values  can  be  assigned,  in  general.  Indeed  if  we  solve  the  preceding 
equation  for  12,  we  get 

Cn±  V(C^n^-  4 AWh cos d) 


12  = 


2  A  cos  6 


from  which  it  is  plain  that  in  general  there  are  two  possible  velocities  of  pre- 
cession for  a  given  gyrostat,  spin  00,  and  inclination  6.  But  if  C~n"  =  4  AWh 
cos  6,  then  there  is  only  one  value  of  12;  and  if  CV'  <  4  AWh  cos  d,  then 
12  is  imaginary,  and  the  gyrostat  will  not  precess  under  the  conditions  imposed. 


2QO  Chap,  xin 

A  gyrostat  whose  center  of  gravity  is  at  the  pivot  will  precess  steadily  for 
certain  conditions  of  impressed  spin,  precession,  and  obliquity.  For,  suppose 
that  spin,  precession,  and  obliquity  are  so  arranged  that  Cn  =  A^cosd; 
then  Ts  =  o,  that  is  no  torque  is  required  to  maintain  the  precession.  Hence 
the  gyrostat,  with  center  of  gravity  at  the  pivot,  would  continue  to  precess. 

It  may  be  instructive  to  consider  the  follov/ing  alternative  analysis.  Let 
hi,  hi,  and  h  be  the  angular  momentums  about  OA,  OB,  and  OC  respectively; 
and  hx,  hy,  and  h  those  about  OX,  OY,  and  OZ.    Then 

hx  =  {h  cos  6  +  hs  sin  9)  cos  xf/, 
hy  =  {hi  cos  d  +  hs  sin  6)  sin  ^, 
and  h  =  —hi  sin- d  -\-  h  cos 6. 

Now  hi,  h,  and  hz  are  constant  (see  preceding  page  for  their  values) ;  the  only 
variable  in  the  right-hand  members  is  xl/;  and  since  the  x,  y,  and  z  axes  are 
fixed  in  direction,  the  rates  at  which  h^,  hy,  and  hz  change  are  respectively 

—  (hi  cos  0  +  ^3  sin  d)  Q,  sin  xj/,     (hi  cos  6  +  hs  sin  d)  12  cos  \}/,     and     o. 
Now  consider  the  instant  when  OC  lies  in  the  plane  ZOX;   then  i/'  =  o,  and 
the  rates  are  o,  (hi  cos  6  -\-  h  sin  6)  2,  and  o.     Hence  the  torques  about  OX 
and  OZ  are  zero,  and  that  about  OY  (the  perpendicular  to  the  axes  of  spin 
and  precession)  is 

(/?i cos ^  +  /?3 sin 0)12=  -  A^^sind cosd  -{-  CnQsind. 

This  result  agrees  with  that  arrived  at  by  the  first  method. 

Gyrostat  in  a  Case.  —  The  foregoing  analysis  must  be  modified  for  a  spin- 
ning gyrostat  in  a  frame  or  case  which  does  not  spin  but  merely  precesses 
with  the  axis  of  spin.  Let  y4  and  C  be  moments  of  inertia  of  the  spinning  part 
as  before,  and  A'  and  C  the  corresponding  moments  of  inertia  of  the  case. 
Then  the  angular  momentums  of  the  case  about  the  axes  OA,  OB,  and  OC 
(Fig.  434)  are  respectively 

-^'12  sine,  o,  and  C^cosO. 
These  may  be  added  to  the  earlier  expressions  for  corresponding  momentums 
of  the  spinning  part  to  arrive  at  values  of  the  angular  momentums  of  the  entire 
gyroscope.  Then  substituting  in  equation  (3)  of  Art.  55  as  before,  we  find 
that  the  necessary  torques  about  OA,  OB,  and  OC  for  steady  precession  are 
respectively 

Ti  =  o, 

Ti=  (Cw  +  C  12  cos  0)12  sine-  (^  +  ^')  ^'sin^cose, 

and  Tz  =  o, 

where  w  =  w  -f  12  cos  6  as  before. 

§  2.  Unsteady  Oblique  Precession.  —  Imagine  a  gyrostat  to  have 
been  started  spinning  in  some  way,  and  then  released  and  left  to  itself  on  a 
frictionless  pivot  under  the  action  of  the  pivot  reaction  and  gravity.  The 
subsequent  motion  will  now  be  investigated. 

Let  Wo  =  the  angular  velocity  of  spin,  and  0o  =  the  angle  between  the  axis 


Ar;  ^6  291 

of  spin  and  the  vertical  at  release.  Let  Fig.  434  represent  the  gyrostat  at 
some  instant  after  its  release;  w  =  the  velocity  of  spin  (velocity  of  the  gyro- 
stat relative  to  the  plane  ZOC) ;  Q  =  the  velocity  of  turning  of  the  plane  ZOC 
(which  we  will  continue  to  call  velocity  of  precession);  and  d  =  the  angle 
ZOC.     We  do  not  assume  w,  fi,  and  d  to  be  constants. 

At  the  instant  of  release,  when  there  is  not  yet  any  motion  of  the  axes  OABC, 
the  total  angular  velocity  of  the  gyrostat  is  coo.  At  a  later  instant  the  angular 
velocity  of  the  gyrostat  is  the  resultant  of  its  velocity  of  spin  00  (relative  to 
the  frame  OABC)  and  the  angular  velocity  of  the  frame.  Now  this  latter 
velocity  has  the  following  components  along  OA,  OB,  and  OC  respectively, 

Osin^,     d*    and     12  cos  0; 

hence  the  (resultant)  angular  velocity  of  the  gyrostat  has  the  following  com- 
ponents along  OA ,  OB,  and  OC  respectively, 

—  12  sin  6,     6,     and    cu  +  12  cos  9  =  n, 

where  n  is  an  abbreviation  for  00  +  cos  6  as  in  §1.  And  the  angular  mo- 
mentums  about  those  same  lines  are 

—  Ailsind,    Bd,    and    Cn. 

According  to  equation  (3),  Art.  55,  the  rate  at  which  the  angular  momentum 
about  OC  is  changing  is 

Cn-^A9. sin d  -  6  -  Bd^smd; 

but  this  rate  equals  zero  since  there  is  no  torque  about  OC.  And  because 
A  =  B,  Cn  =  o;  hence  n  is  constant,  and  therefore  always  equals  its  initial 
value,  that  is, 

n  —  o. 

This  does  not  mean  that  the  spin  velocity,  co,  is  constant. 

Since  there  is  no  torque  about  the  (fixed)  axis  OZ,  the  angular  momentum 
about  that  line  remains  constant;  thus  that  angular  momentum  at  any  instant 
equals  its  initial  value,  or 

A  12  sin  6'  X  sin  0  +  Ccoo  cos  6  =  Cojq  cos  60. 
This  equation  shows  that 

12  =  Ccoo  (cos  do  -  cos  d)~  A  sin2  d,  (i) 

from  which  one  may  compute  the  velocity  of  the  plane  ZOC,  or  the  velocity  of 
precession. 

Investigation  of  the  (nutational)  motion  of  the  spin-axis  in  the  (azimuthal) 
plane  ZOC  can  be  made  simplest  by  means  of  the  princii)le  of  work  and  kinetic 
energy  (Art.  43).  From  the  instant  of  release  of  the  gyrostat  to  any  subse- 
quent instant,  gravity  does  an  amount  of  work  on  the  gyrostat  equal  to  the 
product  of  the  weight  and  the  vertical  descent  of  the  center  of  gravity.     If, 

*  According  to  this  ifluxional)  notation,  a  symbol  with  a  dot  over  it  means  the  time  rate 
of  the  quantity  represented  by  the  symbol;  thus  d  means  dd/dl,  s  means  ds/dl,  etc. 


292 


Chap,  xril 


as  in  §  I,  PF  =  the  weight,  and  h  =  the  distance  from  the  pivot  to  the  center 
of  gravity,  then  the  work  done  by  gravity  is 

Wh  {cos  da  -  cos  e). 

The  initial  kinetic  energy  of  the  gyrostat  is  \  Coo<?,  and  its  kinetic  energy  at  a 
later  instant  is 

Now  the  change  in  kinetic  energy  is  due  to  the  work  done  by  gravity;  hence, 
since  A  =  B, 

i  A  (f22  sin2  d^d'^)  =  Wh  (cos  6^  -  cos  d).  (2) 

From  this  equation  and  (i),  it  is  possible  to  compute  the  angular  velocity 
of  nutation  for  any  value  of  6.  Thus  if  we  eliminate  S2  between  equations 
(i)  and  (2),  we  get 

•     2Wh  .        '      .-     cw  (cos  d,  -  cos  ey 

02  =  — ^  (COS0O  -  cos^j AHmH ^^' 

Now  the  angular  velocity  d  must  be  real,  and  hence  the  right-hand  member 
of  (3)  cannot  be  negative;  it  may  equal  zero  or  be  positive.  The  right-hand 
member  is  zero  (and  then  d  =  0),  when  6  —  do;   also  when  d  equals  di,  where 

cos  01  =  \—  Vi  —  2  X  cos  00  +  X^, 

X  being  an  abbreviation  for  CW/2  WhA.  Any  value  of  d  between  do  and  di 
makes  the  right-hand  member  of  (3)  positive,  and  gives  two  equal  values  of  d 
of  opposite  sign.  Hence  the  spin-axis  oscillates  in  the 
azimuthal  plane,  d  varying  between  do  and  di. 

Whenever  d  is  greater  than  do,  ^  is  positive  (see  equation 
i);  but  when  d  =  do,  then  Q,  =  o.  Hence  the  azimuthal 
plane  rotates  always  in  one  direction  but  its  velocity  is  zero 
every  time  when  the  center  of  gravity  of  the  gyrostat  gets 
into  its  highest  position  {d  =  do). 

In  Fig.  435  the  curve  CoCiCoCi  represents  the  path  of  a 
Fig.  435  point  on  the  spin-axis  of  the  gyrostat;  CoCo  are  the  highest 

and  CiCi  the  lowest  positions  reached  by  the  point;  ZOCo  =  do  and  ZOCi  =  di. 

57.    Principal  Moments  of  Inertia  and  Axes 

§  I.  Moment  of  Inertia  and  Radius  of  Gyration  of  a  body  with  respect 
to  (or  about)  a  line  are  defined  in  Art.  36.  It  is  shown  there,  among  other 
things,  that  the  moments  of  inertia  and  radiuses  of  gyration  of  a  body  with 
respect  to  parallel  lines  are  very  simply  related  (§  2);  and  of  such  moments 
of  inertia  and  radiuses  of  gyration,  the  one  about  the  line  through  the  mass- 
center  is  least.  We  will  now  examine  the  moments  of  inertia  of  a  body  about 
all  lines  through  any  point  of  it.  It  will  be  shown  that  in  general  there  is 
one  Une  about  which  the  moment  of  inertia  is  maximum,  and  a  second  line, 
perpendicular  to  the  first,  about  which  the  moment  of  inertia  is  a  minimum. 


Art.  57  293 

These  two  lines  and  the  one  perpendicular  to  their  plane  at  the  point  in 
question  are  called  the  principal  axes  at  the  point,  and  the  moments  of  inertia 
about  those  lines  are  the  principal  moments  of  inertia  at  the  point.  These 
axes  are  important  dynamically  (see  §  2). 

Let  P  (Fig.  436)  be  a  point  of  a  body  (not  otherwise  shown) ;  0  any  other 
point,  not  in  the  body  necessarily;  and  OA  any  line 
through  0.  Let  X,  /x,  and  v  =  the  direction  cosines 
of  OA  yviih  respect  to  any  coordinate  axes  with  origin 
at  0,  m  =  mass  oi  P,r  =  distance  of  P  from  OA ,  and 
/  =  the  moment  of  inertia  of  the  body  about  OA. 
According  to  definition,  I  =  Smr^.  Now  r^  =  (OPf  - 
(ORy.  OP  is  a  diagonal  of  a  parallelopiped  of  which 
lines  X,  y,  and  s  are  three  intersecting  edges;   hence  ^^^      ^ 

Qp2  =  ^.2  -f  ^2  _|_  22      Q]^  is  one  side  of   the  closed  '  "^^ 

(gauche)  polygon  OZQPRO:  and  since  any  side  of  a  closed  polygon  equals 
the  algebraic  sum  of  the  projections  of  the  other  sides  upon  it,  OR  =  \x  + 
/xy  +  j'S  +  o.     Hence 

y2  =  ^  +  y2  _j.  22  +  (\x  +  juy  +  vzf. 
Ex-panding  this  expression  for  r^  and  arranging  terms  we  would  find  that 

I      /  =  2m[X2(/  +  2^)  +  /x2(22  +  ^2)  _j_  ^2(^2  -f  3,2)  -   2  fiVVZ  -   2  v\ZX  -   2  X/XXy]. 

In  this  (space)  summation  X,  m,  and  v  are  constants;  hence 

7  =  X22m(/ +  22)  +  ....  -  2  iJiv'^jnyz  -   .... 
Now  y2  +  22,  32  +  X-,  and  x"^  +  /  respectively  =  the  squares  of  the  distances 
of  P  from  the  .v,  y,  and  z  axes;  hence  ii  A,  B,  and  C  =  the  moments  of  inertia 
of  the  body  with  respect  to  the  x,  y,  and  z  axes,  we  have 

^=2m(/  +  s2),    5  =  2w(s2  +  .t2),     and    C  ^^mix"" -\- y^). 
The  remaining  summations  in  the  foregoing  expression  for  /  are  the  so-called 
products  of  inertia  of  the  body  with  respect  to  the  two  coordinate  planes  inter- 
secting in  the  x,  y,  and  z  axes  respectively.    Let  D,  E,  and  F  respectively 
denote  these  products  of  inertia,  that  is 

D  =  Jlmyz,    E  =  I^mzx,    and    F  =  I,mxy. 
Then  we  have 

I  =  \-A+  ix^B  +v^C-  2  (jLvD  -  2v\E-  2  X/xF.  (i) 

If  we  know  the  moments  of  inertia  (A,  B,  and  C)  of  a  body  about  each  one  of 
a  set  of  coordinate  axes,  and  the  products  of  inertia  (D,  E,  and  F)  with  respect 
to  each  pair  of  the  coordinate  planes,  then  by  means  of  formula  (i)  we  can  find 
the  moment  of  inertia  /  of  the  body  about  any  line  through  the  origin  of  coor- 
dinates. And  by  means  of  formula  (4)  of  Art.  36  we  can  transfer  this  I  to 
any  parallel  axis  desired.  Thus  the  two  formulas  enable  one  to  "transfer" 
from  the  coordinate  axes  to  any  line  whatsoever. 

Imagine  a  length  OS  laid  off  on  OA  (Fig.  436)  so  that  OS,  which  we  will 
call  p,  is  inversely  proportional  to  the  radius  of  gyration  of  the  body  about 


294 


Chap,  xiii 


OA.  That  is,  ii  k  =  the  radius  of  gyration  and  K  a  factor  of  proportionality, 
then  p  =  K/k.  Such  points  5  for  all  lines  OA  would  lie  on  the  surface  of  an 
ellipsoid  (proved  presently)  called  the  momental  ellipsoid  of  the  body  for  the 
selected  point  0.  Let  X,  Y,  and  Z  =  the  coordinates  of  S;  they  equal  pX, 
p/i,  and  pv  respectively.     Then  equation  (i)  multipUed  by  p^  reduces  to 

AX''  +  BY^  +  CZ2  -  2  DZY  -  2  EYX  -  2  FXY  =  K^M,  (2) 

where  M  =  the  mass  of  the  body.  This  is  the  equation  of  an  ellipsoid  with 
center  at  0  (see  any  standard  work  on  Analytic  Geometry). 

In  general,  the  axes  of  an  ellipsoid  are  unequal  in  length.  Hence  the 
radius  of  gyration  (and  the  moment  of  inertia)  about  the  shortest  axis  of  the 
momental  ellipsoid  is  greater  than  the  radius  of  gyration  (and  moment  of 
inertia)  about  any  other  line  through  the  center  of  the  ellipsoid,  and  the  mo- 
ment of  inertia  about  the  longest  axis  is  less  than  that  about  any  other  line 
through  the  center.  Thus  we  have  shown  that  there  are  two  lines  at  right 
angles  to  each  other  through  any  point  of  a  body  (or  of  its  extension)  about 
which  the  moments  of  inertia  of  the  body  are  maximum  and  minimum.  The 
momental  ellipsoid  might  of  course  be  one  of  revolution  in  a  special  case, 
or  even  a  sphere. 

If  two  of  the  products  of  inertia  equal  zero,  say  E  and  F,  then  the  equation 
of  the  momental  eUipsoid  is 

AX^  +  5F2  +  CZ2  -  2  DYZ  =  Km, 

which  shows  that  the  ellipsoid  is  symmetrical  with  respect  to  the  yz  plane. 

Hence  the  x  axis  coincides  with  one  of  the  axes  of  the  ellipsoid,  that  is  with 

one  of  the  principal  axes  of  the  body  at  the  point  O.     If  the  three  products 

of  inertia  equal  zero,  then  the  ellipsoid  is  symmetrical  with  respect  to  the  three 

coordinate  planes,  and  hence  each  coordinate  axis  is  a  principal  axis  at  the 

origin.     Then  if  h,  h,  and  h  denote  the  principal  moments  of  inertia,  formula 

(i)  becomes 

/  =  X2/i  +  m'/2  +  V'h. 

Symmetrical  Bodies.  —  If  a  homogeneous  body  is  symmetrical  with  respect 
to  a  plane,  then  any  line  perpendicular  to  the  plane  is  a  principal  axis  at  the 
point  where  it  pierces  the  plane.  For,  take  such  line  as  the  x  axis,  and  the  y 
•and  z  axes  in  the  plane.  Then  for  every  particle  of  the  body  whose  coordi- 
nates are  a,  6,  and  c,  there  is  another  one  whose  coordinates  are  —  a,  b,  and 
c;  hence  Hmzx  and  Hmxy  =  o,  and  therefore  as  explained  the  x  axis  is  a 
principal  axis  at  the  origin  of  coordinates. 

If  a  homogeneous  body  has  two  planes  of  symmetry  at  right  angles  to  each 
other,  then  their  intersection  is  a  principal  axis  at  every  point  of  that  line. 
For  if  the  two  planes  be  taken  as  coordinate  planes  and  any  plane  perpendic- 
ular to  them  as  the  third  coordinate  plane,  then  it  is  obvious  that  the  three 
products  of  inertia  equal  zero;  hence  the  intersection  of  the  planes  of  sym- 
metry (one  of  the  coordinate  axes)  is  a  principal  axis  at  the  origin  of  coordi- 
nates (taken  at  any  point  on  the  intersection). 


Art.  57 


295 


Fig.   437 


§  2.  Free  Axes.  —  The  axes  of  principal  moments  of  inertia  at  the  mass- 
center  of  a  rigid  body  are  called  free  axes  of  the  body  because  they  possess 
a  certain  property  which  may  be  described  as  follows:  If  the  body  could  be 
set  to  rotating  about  any  one  of  these  axes  and  then  left  to  itself  entirely  free 
from  all  external  forces,  even  gravity,  it  would  continue  to  rotate  about  that 
axis.  To  demonstrate  this,  we  will  imagine  this  axis  to  be  a  shaft  resting  in 
bearings,  and  then  show  that  the  bearings  would  exert  no  pressures  whatever 
on  the  shaft.  It  will  follow  that  such  bearings  are  not  necessary  to  hold  the 
shaft  in  position.  Let  B1B2  (Fig.  437)  be  the  axis,  Bi  and  B^  the  bearings, 
0  the  mass-center  of  the  body,  and  co  =  the  angular  velocity.  Also  let  O  be 
the  origin  of  the  axes  x,  y,  and  z  as  shown,  P  and  Q  -=  the  reactions  of  the 
bearings  on  the  shaft;  and  P^,  Py,  and  Pz  =  the  axial  components  of  P,  and 
Qx,  Qy,  and  Qz  =  those  of  Q.  Evidently,  Pz  or  Qz  =  o.  Since  the  mass- 
center  is  at  rest,  the  sums  of  the  x,  y,  and  z 
components  of  all  the  forces  (P  and  Q)  acting 
on  the  body  equal  zero.     Hence 

Pg  =  Q^  =  o]  Px  and  Qx  are  opposite; 
also  Py  and  Qy. 
That  is,  the  external  forces  consist  of  two 
couples,  Px  and  Qx,  and  Py  and  Qy.  Let  A 
be  one  of  the  particles  of  the  body;  m  =  its 
mass;  r  =  its  distance  from  the  axis;  d  == 
the  (varying)  angle  which  r  makes  with  xz  plane  (as  shown);  x,  y,  and 
2  =  the  coordinates  of  A ;  z'x,  Vy,  and  v^  =  the  axial  components  of  the  veloc- 
ity of  A.  Since  v^  =  o,  the  angular  momentums  of  A  about  the  x  and  y 
axes  are  respectively 

—  mVyZ  =  —  mv cos 6  •  z  =  —  mroi cos d  -z  =  —  oomxz, 
and  mVxZ  =  —  mvsind  '  z  =  —  mru  sin 0  •  z  =  —  wmyz. 

Hence  the  angular  momentums  of  the  entire  body  about  the  x  and  y  axes 

respectively  are 

—  w^mxz    and     —  coSmyz. 

Since  the  z  axis  is  a  principal  axis,  these  summations  (or  products  of  inertia) 
equal  zero;  that  is  the  angular  momentums  about  the  x  and  y  axes  equal  zero, 
at  all  times.  It  follows  that  there  is  no  torque  about  either  axis  at  any  time; 
hence  there  are  no  such  couples,  PxQx  and  PyQy 

If  the  axis  of  rotation  ByBi  is  not  a  principal  axis,  then  the  angular  momen- 
tums —  oiZmxz  and  —  wSwyz  are  not  always  zero  nor  are  they  constant  in 
value  during  a  revolution;  hence  the  torques  of  the  couples  PxQx  and  PyQy 
are  not  always  zero.  Such  couples  can  be  sensed  roughly  by  supporting  an 
irregular  shaped  body  by  means  of  one's  hands  as  bearings  and  then  making 
it  rotate.  Of  course  one  would  feel  the  dead  weight  of  the  body  but  also 
pulls  and  pushes  due  to  the  tendency  or  effort  of  the  axis  of  rotation  to  get 
away.  If  a  regular  body  were  selected,  one  would  be  apt  to  rotate  it  about 
a  principal  axis,  and  so  miss  the  effect  just  described. 


2q6  Chap,  xiii 

In  general  a  body  has  three  free  axes,  the  axes  of  greatest,  least,  and  mean 
moments  of  inertia  —  axes  through  the  mass-center  being  understood.  There 
is  an  interesting  difference  among  these  axes,  namely,  rotation  about  the  axis 
of  greatest  or  least  moment  of  inertia  is  stable,  while  rotation  about  the  axis 
of  mean  moment  is  unstable.  That  is,  if  the  body  were  set  rotating  about 
either  of  the  first  two  axes,  then  any  slight  deviation  of  the  axis  of  rotation 
from  such  principal  axis  would  not  be  followed  by  continually  increasing 
deviations;  but  if  the  initial  rotation  be  about  the  axis  of  mean  moment  then 
slight  deviation  is  followed  by  still  greater  change.  Explanation  of  these 
properties  would  take  us  too  far  afield. 

58.   Any  Motion  of  a  Rigid  Body;   Summary  of  D3niamics 

§  I.  Any  Motion  or  a  Rigid  Body.  —  A  rigid  body  can  be  displaced 
from  one  position  A  into  another  position  B  by  means  of  a  translation  followed 
by  a  rotation.  For,  it  is  obvious  that  a  translation  can  be  selected  so  as  to 
move  any  chosen  point  0  of  the  body  from  its  original  position  (in  A)  into  its 
final  position  (in  B).  From  this  intermediate  position,  the  body  can  be  put 
into  its  final  position  by  means  of  a  rotation  (of  suitable  amount)  about  a 
(certain)  fixed  axis  through  the  final  position  of  0  (see  Art.  54).  The  displace- 
ment might  be  effected  in  the  reverse  order,  that  is  a  rotation  followed  by  a 
translation.  For,  a  rotation  about  a  fixed  line  through  the  "base  point  "  0 
could  be  made  so  as  to  put  the  two  lines  OP  and  OQ,  P  and  Q  being  two  points 
of  the  body  not  in  line  with  0,  parallel  to  their  final  positions  (in  B)\  and  a 
suitable  translation  would  put  those  lines  (and  the  body)  into  their  final 
positions. 

Evidently,  the  rotation  and  the  translation  could  be  made  simultaneously. 
Therefore  any  actual  motion  of  a  body  from  one  position  into  another  may 
be  regarded  as  a  succession  of  infinitesimal  simultaneous  translations  and 
rotations.  All  the  translations  may  refer  to  the  same  base  point,  but  in  gen- 
eral the  successive  rotations  do  not  occur  about  the  same  line  of  the  body. 
Thus  we  may  regard  any  motion  of  a  rigid  body  as  consisting  of  a  translation 
(in  which  each  point  of  the  body  moves  just  like  the  base  point),  combined  with 
a  rotation  about  a  line  through  the  base  point,  the  hne  shifting  about  in  the 
body,  generally.  There  is  not  only  such  (kinematic)  independence  of  trans- 
lation and  rotation  as  just  explained,  but  if  we  take  the  center  of  gravity 
as  base  point,  then  there  is  also  independence  of  translation  and  rotation 
dynamically.  That  is  to  say,  we  may  ascertain  the  translation  (or  motion 
of  the  center  of  gravity)  quite  independently  of  the  rotation;  and  the  rota- 
tion about  the  center  of  gravity  quite  independently  of  the  translation.  We 
proceed  to  demonstrate  this  independence. 

As  already  shown  in  Art.  34,  the  acceleration  of  the  mass-center  of  a  body 
(even  if  not  rigid)  may  be  determined  as  though  all  the  material  of  the  body 
were  concentrated  at  the  mass-center,  and  all  the  external  forces  were  applied 
at  that  (dense)  point.    Therefore  if  Rx,  Ry,  and  Rz  denote  the  algebraic  sums 


Art.  58  297 

of  the  components  along  x,  y,  and  2  axes  respectively,  of  all  the  external 
forces  acting  on  a  body;  a^,  Oy,  and  Cj  the  components  of  the  acceleration  of 
its  mass-center,  and  M  its  mass;   then 

R:c  =  MUj:,      Ry  =  May,      Rz  =  MUz. 

And,  in  Art.  46  it  was  shown  that  the  component  along  any  line,  of  the  linear 
momentum  of  a  moving  body  is  just  the  same  as  though  the  material  of  the 
body  were  concentrated  at  the  (moving)  mass-center;  therefore  if  Vx,  Vy,  and 
Vz  denote  the  axial  components  of  the  velocity  of  the  mass-center,  then 

Rx=    ^^  AIV„         Ry    =    -  MVy,         K^=    Jf  Mvz. 

To  see  that  the  rotation  about  the  mass-center  is  independent  of  the  trans- 
lation, let  us  apply  the  principle  that  the  torque  of  all  the  external  forces  about 
any  line  equals  the  rate  at  which  the  angular  mo- 
mentum  about  that  line  is  changing  (Art.  48),  taking 
the  line  through  the  mass-center.  Let  O  (Fig.  438) 
be  the  mass-center,  and  P  any  other  point  of  the  mov- 
ing body;  a,  b,  and  c  =  the  (changing)  coordinates 
of  0  with  respect  to  fixed  axes  QX,  QY ,  and  QZ]  x,  y, 
and  2  =  the  coordinates  of  P  with  respect  to  the  same 
axes,  and  x' ,  y',  and  z'  =  the  coordinates  of  P  with 
respect  to  a  parallel  set  of  axes,  the  origin  being  at  0. 

Furthermore,  let  m  =  mass  of  P,  and  t'.,  Vy,  and  Vz  =  the  x,  y,  and  2  compo- 
nents of  the  velocity  of  P;  then  the  angular  momentum  of  P  about  the  z' 
axis  say  (through  O  and  parallel  to  the  2  axis)  =  mvyx'  —  mv^y' ,  and  the  angu- 
lar momentum  of  the  entire  body  about  that  Une  equals 

Umi^Pyx'  —  Vxy'). 

Now  X  =  x'  -\-  a,  and  y  =  y'  -\-  b;  hence  Vx  =  dx'/dt  +  da/dt  and  Vy  =  dy'/dt 
-\-  db/dt.     Therefore,  the  angular  momentum  = 

Sw(.r'  dy'/di  +  x'  db/dt  -  y'  dx'/dt  -  y'  da/dt)  = 
Hmx'  dy'/dt  +  (db/dt)  Jlmx'  -  ^my'  dx'/dt  -  {da/dl)  Y.my' . 

Since  Swx'  and  Hmy'  =  o,  the  second  and  fourth  terms  equal  zero.  Hence 
the  angular  momentum  equals 

.     llmix'  dy'/dt  -  y'  dx'/dt) . 

Now  this  expression  does  not  depend  on  the  motion  of  the  center  of  gravity 
at  all;  moreover,  it  is  just  like  the  expression  for  the  angular  momentum 
of  a  body  rotating  about  a  fixed  point,  with  respect  to  a  line  through  that 
point  (see  Art.  55).  Therefore  if  Tx,  Ty,  and  Tz  denote  the  torques  (mo- 
ment sums)  of  the  external  forces  acting  on  a  body  about  any  lines  .v,  y,  and 
2  through  the  mass-center,  and  hx,  hy,  and  hg  the  angular  momentums  of  the 
body  about  those  lines,  we  may  write 

T    -—i       r    -  ^       T   =^ 
^'~  dt'       '       dt'       '       dt' 

and  use  expressions  for  hx,  hy,  and  hz  and  their  rates  given  in  Art.  55. 


298  Chap,  xni 

To  further  emphasize  the  independence  of  motion  of  the  mass-center  and 
the  motion  of  the  body  ^bout  that  point,  let  us  consider  the  kinetic  energy  of 
a  body  having  any  motion.  Let  0  (Fig.  438)  be  the  center  of  mass  of  the  body. 
We  will  regard  the  motion  as  consisting  of  a  translation  like  the  motion  of  0, 
and  a  rotation  about  some  line  through  O.  Let  Vi,  Vy,  and  Vz  be  the  axial 
components  of  the  velocity  of  0;  Ux-  (^y,  and  Uz  the  axial  components  of  the 
angular  velocity  about  0.  Then  the  components  of  the  velocity  of  any  par- 
ticle P  are  (see  Art.  54) 

Vj;  =  Vx-{-  z'iCy  —  y'wz, 

Vy=Vy-\-  x'Uz  —  z'cOx, 
Vz  ^   Vz-\-  j'cOx  —  .v'cOy. 

Since  the  kinetic  energy  of  the  particle  is  h  mv^  =  \m  (vx'  +  Vy^  +  vj^),  the 
kinetic  energy  of  the  body  is  2  ^  mv"^,  or 

I  M(Vx^  +  Vy-  +  Vz^)-\-  \  IjfJix    +   2  ^\f^y    +   2  ^if^z     —  Japi^z  —  /yWzOJx  —  J Jji^y. 

Now  the  first  term  of  this  expression  equals  \  Mv^,  and  is  the  value  which 
the  kinetic  energy  would  have  if  all  the  material  were  concentrated  at  the 
mass-center  and  were  moving  with  it.  The  remainder  of  the  expression  is 
the  value  which  the  kinetic  energy  of  the  body  would  have  if  the  center  of  mass 
were  fixed  (Art.  55). 

Once  more,  let  us  consider  the  angular  momentum  of  a  body  with  respect 
to  a  fixed  line,  say  the  z  axis  of  Fig.  438.  From  the  definition  (Art.  48)  the 
angular  momentum  about  the  s  axis  of  the  particle  P  is 

mivyX  —  Vxy)  =  ni[{vy-]-  x'iOz  —  z'wx)  -v  —  (^x  +  s'ojy  —  y'w^y], 

and  the  angular  momentum  of  the  body  equals  the  sum  of  all  such  expressions 
as  the  last,  or 

M(VyX  —   V^)   —  JyUix  —  Jii^y  +  Jz^z- 

Now  the  first  term  is  the  value  which  the  angular  momentum  would  have  if 
all  the  material  were  concentrated  at  the  center  of  mass  and  moving  with  it; 
the  remainder  of  the  expression  is  the  value  which  the  angular  momentum 
would  have  if  the  mass-center  were  fixed  (see  Art.  55). 

§2.  Summary;  Motion  of  a  Rigid  Body.  —  The  following  summary 
is  based  upon  the  order  of  development  of  dynamics  followed  in  this  book; 
it  may  therefore  be  regarded  as  a  brief  review. 

Motion  of  Translation  (rectilinear  or  otherwise).  — -  The  resultant  of  all  the 
external  forces  acting  on  the  body  is  a  single  force.  Its  line  of  action  passes 
through  the  mass-center,  and  hence  the  external  forces  have  no  torque  about 
any  line  through  that  point.  The  resultant  and  the  acceleration  of  the  mov- 
ing body  have  the  same  direction ;  hence  the  algebraic  sum  of  the  components 
of  the  external  forces  at  right  angles  to  the  acceleration  equals  zero.  The 
acceleration  is  proportional  to  the  resultant  directly  and  to  the  mass  of  the 
body  inversely;  or  a  =  R/M  (where  a  =  acceleration,  R  =  resultant,  and 
M  =  mass)  if  systematic  units  be  used.     The  (linear)  momentum  of  the  body 


Art.  58  299 

=  Mv  =  (W/g)v,  where  v  =  velocity,  W  =  weight,  and  g  =  acceleration  due 
to  gravity.     The  kinetic  energy  of  the  body  =  |  Mv^  =  |  (W/g)  v^. 

Rotation  about  a  Fixed  Axis.  —  The  torque  of  all  the  external  forces  about 
the  axis  of  rotation  and  the  angular  acceleration  of  the  rotating  body  are  alike 
in  sense.  The  angular  acceleration  is  proportional  to  the  torque  directly 
and  to  the  moment  of  inertia  of  the  body  (with  respect  to  the  axis  of  rotation) 
inversely;  or  a  =  T/I  (where  a  =  angular  acceleration,  T  =  torque,  and 
/  =  moment  of  inertia),  if  systematic  units  be  used.  (It  is  generally  con- 
venient to  take  I  =  Mk^  =  (W/g)  h^  where  k  =  radius  of  gyration  of  the  body 
about  the  axis  of  rotation.)     We  have  also 

SF„  =  Man,     ^Ft  =  Mat,    and     2F3  =  o. 

The  summations  mean  the  algebraic  sums  of  the  components  of  the  external 
forces  —  including  the  axle  reaction  if  any  —  along  three  certain  lines,  namely, 
—  (i)  the  perpendicular  to  the  axis  of  rotation  through  the  mass-center,  (2) 
the  perpendicular  to  the  plane  of  the  line  just  mentioned  and  the  axis  of  rota- 
tion, and  (3)  the  axis  of  rotation.  Symbols  a„  and  at  denote  the  components 
of  the  acceleration  of  the  mass-center  along  the  first  two  lines  respectively; 
an  =  ro)^  and  at  =  ra,  where  r  =  the  distance  from  the  mass-center  to  the 
axis  of  rotation,  co  =  the  angular  velocity,  and  a  =  the  angular  acceleration 
of  the  body.  If  the  angular  velocity  is  constant  then  I1F(  =  o;  if  the  mass- 
center  is  in  the  axis  of  rotation,  then  the  three  summations  equal  zero. 

The  angular  momentum  of  the  body  =  loo  —  Mk^oj  =  (W/g)  k^co;  its  kinetic 
energy  =  i  /co^  =  i  MyfeW  =  i  {W/g)  kW. 

Uniplanar  Motion.  —  It  may  be  regarded  as  a  combined  translation  and 
rotation.     Motion  of  the  mass-center  is  given  by 

SFx  =  Max    and    llFy  =  May, 

where  SF^  and  HFy  respectively  mean  the  algebraic  sums  of  the  components 
of  the  external  forces  along  axes  x  and  y  in  the  plane  of  the  motion;  and  a^ 
and  ay  =  the  x  and  y  components  of  the  acceleration  of  the  mass-center.  The 
rotation  of  the  body  about  the  mass-center  is  given  by  T  =  la,  where  T  = 
the  torque  of  the  external  forces  about  the  perpendicular  to  the  plane  of  the 
motion  through  the  mass-center,  /  =  the  moment  of  inertia  of  the  body  about 
that  line,  and  a  =  the  angular  acceleration  of  the  body.  The  kinetic  energy 
of  the  body  is  given  by  \  Mv^  +  |  loo^,  where  v  =  the  velocity  of  the  mass- 
center  and  CO  =  the  angular  velocity  of  the  body  at  the  instant  in  question. 

Rotation  about  a  Fixed  Point.  —  The  principle  of  motion  of  the  mass-center 
(§  3)  furnishes  three  independent  equations  of  motion  of  the  mass-center  like 

2Fx  =  Md„ 

where  the  symbols  have  meanings  already  explained.  The  axes  x,  y,  and  z 
must  not  be  parallel.  The  principle  of  torque  and  angular  momentum  fur- 
nishes three  independent  equations  like 

Tx  =  dhx/dt, 


300 


Chap,  xiii 


where  Tx  is  the  torque  of  the  forces  acting  on  the  body  about  any  line  x,  and  //^ 
is  the  angular  momentum  of  the  body  about  the  same  line.  That  line  and 
the  other  two  may  be  taken  at  pleasure;  generally  hnes  through  the  fixed 
point  are  simplest. 

Any  {Solid)  Motion.  —  Motion  of  the  mass-center  and  rotation  about  the 
mass-center  are  independent  (§  i).  We  may  treat  these  motions  separately; 
the  first  as  the  motion  of  a  particle  whose  mass  equals  that  of  the  body  under 
the  action  of  forces  Uke  those  acting  on  the  body;  the  second  as  though  the 
body  were  fixed  at  the  mass-center. 

§  3.   Summary;  Motion  of  Any  System  of  Particles,  Solid  or  Fluid. 

—  We  will  call  the  system  a  body  but  without  implying  that  it  is  rigid  except 
as  noted. 

Principle  of  Motion  of  Mass-Center.  —  The  motion  of  the  mass-center  does 
not  depend  at  all  on  the  internal  forces;  it  moves  just  as  if  all  the  material 
of  the  moving  body  were  concentrated  at  the  mass-center  and  all  the  external 
forces  were  applied  to  that  (dense)  point  in  their  actual  directions  of  course. 
Thus  the  component  of  the  acceleration  of  the  mass-center  along  any  line 
is  proportional  to  the  algebraic  sum  of  the  components  of  the  external  forces 
along  that  line  directly  and  to  the  mass  of  the  body  inversely.  Or,  if  sys- 
tematic units  are  used  a^  =  2Fx  -^  M,  where  ax  and  S/^x  =  the  mentioned 
component  acceleration  and  algebraic  sum  respectively  and  M  =  mass.  The 
foregoing  equation  is  generally  written 

2F^  =  Max. 

Principle  of  Force  and  Momentum.  —  The  algebraic  sum  of  the  components 

—  along  any  line  —  of  the  external  forces  acting  on  the  body  at  any  instant 
equals  the  rate  at  which  the  component  (along  the  same  Une)  of  the  momen- 
tum of  the  body  is  changing  them.     Or 

where  Vx  =  the  component  of  the  velocity  of  the  mass-center  along  the  line 
called  X.    Mvx  —  the  x  component  of  the  linear  momentum  of  the  body. 

Principle  of  Impulse  and  Momentum.  —  The  algebraic  sum  of  the  impulses 
of  the  components  —  along  any  line  —  of  all  the  external  forces  for  any  in- 
terval equals  the  increment  in  the  component  of  the  momentum  of  the  body 
along  that  line  for  that  interval.     Or 


^r 


Fxdt  =  Mvx"  -  MvJ. 


Principle  of  Torque.  —  The  torque  of  all  the  external  forces,  acting  on  any 
body,  equals  the  torque  of  the  resultants  of  all  the  forces  acting  on  the  par- 
ticles of  the  body,  all  torques  being  taken  about  any  line.  An  expression 
for  this  latter  torque  was  deduced  in  Art.  48,  the  line  about  which  torque  was 
taken  being  called  a  z  coordinate  axis;  it  is  21w  {ayX  —  axj),  where  m.  =  the 
mass  of  any  particle,  x  and  y  =  the  coordinates  of  the  particle,  and  ax  and 


i: 


Art.  58  301 

Oy  =  the  X  and  y  components  of  its  acceleration  all  at  the  instant  in  question. 
Thus  the  principle  gives 

Tg  =  2w(oyX  —  ax}'), 
where  Tg  means  the  torque  of  all  the  external  forces  about  the  z  axis. 

Principle  of  Torque  and  Angular  Momentum.  —  The  torque  of  the  external 
forces  acting  on  the  moving  body  about  any  line  equals  the  rate  at  which  the 
angular  momentum  of  the  body  about  that  line  is  changing  (Art.  48).     Or 

Tz  =  —'^{mVyX  —  mvjry), 

the  torque  and  angular  momentum  being  taken  about  a  z  axis  of  a  coordinate 
frame.  If,  in  a  given  case,  there  is  no  torque  during  an  interval  then  the 
angular  momentum  about  that  line  remains  constant;  this  is  the  principle  of 
conservation  of  angular  momentum. 

Principle  of  Angular  Impulse  and  Momentum.  —  The  angular  impulse  of  all 
the  external  forces  —  about  any  line  —  for  any  interval  equals  the  increment 
in  the  angular  momentum  of  the  body  about  that  line  and  for  that  interval. 
Or 

Tzdt  =  AH  (mVyX  —  mv^y). 
'i' 

Principle  of  Work  and  Kinetic  Energy.  —  The  total  work  done  upon  a  body 
by  all  the  external  and  internal  forces  during  any  displacement  of  the  body 
equals  the  increment  in  the  kinetic  energy  of  the  body  during  the  interval. 
If  the  total  work  is  positive  then  there  is  a  real  gain;  if  negative,  then  there 
is  loss.  If  the  body  is  a  rigid  one,  the  internal  forces  do  no  work;  the  total 
work  done  upon  the  body  by  the  external  forces  equals  the  increment  in  its 
kinetic  energy. 

Principle  of  Conservation  of  Energy.  —  If  a  body  is  isolated  so  that  it  is 
beyond  the  influence  of  other  bodies,  then  during  any  change  of  condition 
of  the  body,  the  amount  of  its  energy  remains  constant.  There  may  be  a 
transfer  of  energy  from  one  part  of  the  body  to  another,  but  the  total  gain 
or  loss  in  one  part  is  exactly  equivalent  to  the  loss  or  gain  in  the  remainder. 

D'Alembert's  Principle,  not  heretofore  discussed.  —  The  resultant  of  all  the 
forces  acting  on  any  particle  of  a  body  is  called  the  effective  force  for  that  par- 
ticle. Its  magnitude  equals  the  product  of  the  mass  and  acceleration  of  the 
particle;  its  direction  is  the  same  as  that  of  the  acceleration.  The  group  of 
effective  forces  for  all  the  particles  of  a  body  is  called  the  effective  system  (of 
forces)  for  the  body.  It  should  be  noted  that  these  forces  are  fictitious  or 
imaginary,  equivalent  respectively  to  the  actual  forces  acting  upon  the  par- 
ticles. The  principle  may  be  stated  in  two  forms:  —  (a)  The  external  system 
of  forces  and  the  effective  system  are  equivalent,  and  (b)  the  external  system 
and  the  reversed  effective  system  jointly  balance,  or  are  in  equilibrium. 


APPENDIX    A 


THEORY  OF  DIMENSIONS  OF  UNITS 

§  I.  Dimensions  of  Units. — The  magnitude  of  a  quantity  is  expressed 
by  stating  how  many  times  larger  it  is  than  a  standard  quantity  of  the  same 
kind  and  naming  the  standard.  Thus,  we  say  that  a  certain  distance  is  lo 
miles,  meaning  that  the  distance  is  lo  times  as  great  as  the  standard  distance, 
the  mile.  The  number  expressing  the  relation  between  the  magnitude  of 
the  quantity  and  the  standard  (the  number  lo  in  the  illustration)  is  called  the 
numeric  (or  numerical  value)  of  the  quantity,  and  the  standard  is  called  the 
unit. 

A  unit  for  measuring  any  kind  of  quantity  may  be  selected  arbitrarily,  but 
it  must  of  course  be  a  quantity  of  the  same  kind  as  the  quantity  to  be  measured. 
Thus,  as  unit  of  velocity  we  might  select  the  velocity  of  light,  as  unit  of  area 
the  area  of  one  face  of  a  silver  dollar,  etc.  Many  units  in  use  are  arbitrarily 
chosen,  that  is  without  reference  to  another  unit  (for  example,  the  bushel  and 
the  degree);  but  generally  it  is  convenient  practically  to  define  them  with 
reference  to  each  other.  All  mechanical  and  nearly  all  physical  quantities 
can  be  defined  in  terms  of  three  arbitrarily  selected  units,  not  dependent  on 
any  other  units.  These  are  called  fundamental  units,  and  the  others,  defined 
with  reference  to  them,  derived  units.  It  is  customary  in  works  on  theoretical 
mechanics  and  physics  to  choose  as  fundamental  the  units  of 

length,  mass,  and  time; 
but  it  is  sometimes  more  convenient  to  take  as  fundamental  the  units  of 

length,  force,  and  time. 

We  give  an  analysis  of  derived  units  with  reference  to  each  of  these  sets  of 
fundamentals,  and  two  tables  in  which  the  absolute  units  are  referred  to  the 
first  set  of  fundamentals  and  the  gravitational  units  to  the  second  set.  But 
either  set  might  serve  as  fundamentals  for  all  absolute  and  gravitational 
units. 

A  statement  of  the  way  in  which  a  derived  unit  depends  on  the  funda- 
mental units  involved  in  it  is  called  a  statement  of  the  dimensions  of  the 
unit.     For  example, 

one  square  yard  _  (one  yard,  or  three  feet)^  __ 
one  square  foot  (one  foot)^ 

302 


APPENDIX   A  3®3 

Thus,  a  unit  of  area  depends  only  on  the  unit  of  length  used,  and  the  unit  of 
area  varies  as  the  square  of  the  unit  of  length.  This  relation  is  expressed  in 
the  form  of  a  "dimensional  equation"  as  follows: 

(unit  area)  =  (unit  length)^, 

and  briefly  a  unit  area  is  said  to  be  "two  dimensions  in  length."  Similarly,  a 
unit  volume  is  said  to  be  three  dimensions  in  length.  We  proceed  to  de- 
termine "dimensional  formulas"  for  the  vmits  of  several  of  the  quantities  of 
mechanics.  The  student  should  be  able  to  determine  formulas  (see  sub- 
sequent tables)  for  the  others. 

Fe/oaVj.  —  According  to  the  definition  of  velocity  (Art.  28),  a  unit  ve- 
locity is  directly  proportional  to  the  unit  length  and  inversely  to  the  unit 
time;  hence  if  V,  L,  and  T  denote  units  of  velocity,  length,  and  time  respec- 
tively, the  dimensional  equation  is 

V=  L/T  =  LT-S 

and  a  unit  velocity  is  one  dimension  in  length  and  minus  one  in  time. 

Acceleration. — According  to  the  definition  of  acceleration  (Art.  28),  a 
unit  acceleration  is  proportional  directly  to  the  unit  velocity  and  inversely  to 
the  unit  time;  hence  if  A  denotes  unit  acceleration,  the  dimensional  equation 
is 

and  a  unit  acceleration  is  one  dimension  in  length  and  minus  two  in  time. 

Angular  Velocity.  —  According  to  the  definition  of  angular  velocity  (Art. 
37),  a  unit  angular  velocity  is  proportional  directly  to  the  unit  angle  and 
inversely  to  the  unit  time;  hence  if  <o  and  6  denote  units  of  angular  velocity 
and  angle  respectively,  the  dimensional  equation  is 

(0  =  6/T    or  (0  =  T-\ 

since  units  of  angle  (degree,  radian,  etc.)  are  independent  of  the  fundamental 
units.     A  unit  angular  velocity  is  therefore  minus  one  dimension  in  time. 

Angular  Acceleration.  —  kccording  to  the  definition  of  angular  velocity 
(Art.  37),  a  unit  angular  acceleration  is  proportional  directly  to  the  unit  an- 
gular velocity  and  inversely  to  the  unit  time;  hence  if  a  denotes  unit  angular 
acceleration,  the  dimensional  equation  is 

a  =  a)/T  =  T-2, 

and  a  unit  angular  acceleration  is  minus  two  dimensions  in  time. 

Force.  —  In  accordance  with  the  equation  of  motion  of  a  particle  (Art.  31), 

R  =  ma,  or 

"force  =  mass  X  acceleration;" 

that  is,  the  unit  force  is  directly  proportional  to  the  units  of  mass  and  accel- 
eration.    Hence  if  F  and  M  denote  units  of  force  and  mass  respectively,  the 

dimensional  equation  is 

F  =  MA  =  LMT-2, 


304 


APPENDIX   A 


ABSOLUTE  SYSTEMS 


Names  of  Quantities. 


Length 

Mass 

Time 

Velocity 

Acceleration 

Angular  velocity 

Angular  acceleration 

Force 

Weight 

Moment  of  mass 

Moment  of  inertia  (body) . 

Moment  of  force 

Work 

Energy 

Power 

Impulse 

Momentum 

Density 

Specific  weight 

Moment  of  area 

Moment  of  inertia  (area). . 

Stress 

Stress  intensity 


Dimen- 
sional 
Formulas. 


Names  of  Units. 


C.G.S. 


L 

M 

T 
LT-i 
LT-2 

T-i 

f— 2 

LMT-2 

LMT-2 

LM 

L2M 

L2MT-2 

L^MT-^ 

L2MT-2 

L=MT-3 

LMT-i 

LMT-i 

L-^M 

L-2MT-1 

U 

L* 

LMT-2 

L-iMT-2 


centimeter  (cm) 

gram  (gr) 

second  (sec) 

cm/sec  ("kine") 

cm/sec^  ("  spoud") 

rad/sec 

rad/sec^ 

dyne 

dyne 

gr-cm 

gr-cqi^ 

cm-dyne 

cm-dyne  ("  erg  ") 

cm-dyne  ("  erg  ") 

erg/sec 
dyne-sec  ("  bole  ") 


dyne-sec  ("  bole 

gr/cm^ 

dyne/cm' 

cm' 

cm* 

dyne 

dyne/cm^ 


') 


F.P.S. 


foot  (ft) 

pound  (lb) 

second  (sec) 

ft/sec 

ft/sec^ 

rad/sec 

rad/sec^ 

poundal  (pdl) 

pdl 

Ib-ft 

Ib-ft^ 

ft-pdl 

ft-pdl 

ft-pdl 

ft-pdl/sec 

pdl-sec 

pdl-sec 

lb/ft' 

pdl/ft» 

ft' 

ft* 

pdl 

pdl/ft2 


and  a  unit  force  is  one  dimension  in  length,  one  in  mass,  and  minus  two  in 
kime. 

Mass.  —  If  we  regard  length,  force,  and  time  as  fundamental  units,  then 
\\ie  last  equation  written  as  follows  is  the  dimensional  equation  for  a  unit 

mass: 

M  =  FTVL  =  L-1FT2, 

and  a  unit  mass  is  minus  one  dimension  in  length,  one  in  force,  and  two  in 

time. 

Work.  —  According  to  the  definition  of  work  (Art.  40),  the  unit  of  work  is 
directly  proportional  to  the  units  of  force  and  length;  hence  if  W  denotes 
unit  work,  the  dimensional  equation  is 

W  =  LF  =  L2MT-2, 

and  a  unit  work  is  one  dimension  in  length,  one  in  force,  or  two  in  length, 
one  in  mass,  and  minus  two  in  time. 

Power.  — According  to  the  definition  of  power  (Art.  42),  a  unit  of  power  is 
proportional  directly  to  the  unit  work  and  inversely  to  the  unit  time;  hence 
if  P  denotes  unit  of  power,  the  dimensional  equation  is 

p  =  w/T  =  LFT-i  =  L^MT-s, 

and  a  unit  power  is  one  dimension  in  length  and  force  and  minus  one  in  time^ 
or  two  in  length,  one  in  mass,  and  minus  three  in  time. 


APPENDIX   A 


305 


GRAVITATION   SYSTEMS 


Names  of  Quantities. 


Length 

Force 

Time 

Velocity 

Acceleration 

Angular  velocity. . .  . 
Angular  acceleration 

Mass 

Weight 

Moment  of  mass.  .  .  . 
Moment  of  inertia.  .  , 
Moment  of  force .... 

Work 

Energy 

Power 

Impulse 

Momentum 

Density 

Specific  weight 

Moment  of  area 

Moment  of  inertia.  . 

Stress 

Stress  intensity 


Names  of  Units. 


P.P.  (force)  S. 


M.K.  (force)  S. 


foot  (ft) 

pound  (lb) 

second  (sec) 

ft/ sec 

ft/sec'^ 

rad/sec 

rad/sec^ 

"  slug  "  (si) 

lb 


meter  (m) 

kilogram  (kg) 

second  (sec) 

m/sec 

m/sec'' 

rad/sec 

rad/sec^ 

metric  slug  " 

kg 


(msl) 


sl-ft 

msl-m 

Sl-ft2 

msl-m^ 

ft-lb 

kg-m 

ft-lb 

kg-m 

ft-lb 

kg-m 

ft-lb/sec 

kg-m/sec 

lb-sec 

kg-sec 

lb-sec 

kg- sec 

sl/ft^ 

msl/m* 

Ib/ft^ 

kg/m^ 

ft» 

m' 

ft^ 

m* 

lb 

kg 

lb/ft2 

kg/m^ 

§  2.  Applications  of  the  Theory  of  Dimensions.  —  A  knowledge  of 
the  theory  of  dimensions  is  probably  of  most  value  to  the  beginner  as  a  help 
to  a  clear  understanding  of  the  different  mechanical  quantities  and  the  rela- 
tions between  them.  The  theory  is  useful  practically  in  other  ways,  three  of 
which  we  mention, 

(i)  As  a  test  of  the  accuracy  of  equations  between  mechanical  quantities.  — 
Such  an  equation  if  rationally  and  correctly  deduced  must  be  homogeneous, 
that  is  the  terms  in  it  must  be  the  same  in  kind.  To  ascertain  whether  terms 
are  the  same  in  kind  we  write  the  dimensional  form  of  the  equation,  reduce  the 
terms  to  their  simplest  forms  and  compare;  if  they  are  aUke,  the  terms  are 
the  same  in  kind.    To  illustrate  let  us  consider  equation  (4),  Art.  25, 


T  =  -iva 
2 


1  + 


C2    Y 


where  T  denotes  tension  (or  force),  w  weight  (or  force)  per  unit  length,  and  a 
and/  lengths.     Using  L,  M,  and  T,  the  dimensional  form  of  the  equation  is 


LMT 


'-2  — 


LMT 


V 


Since  the  right-hand  member  reduces  to  LMT-^,  the  two  members  are  alike 
in  kind,  as  they  should  be.     The  coefficient  |  and  the  term  i  were  omitted 


3o6 


APPENDIX   A 


from  the  dimensional  equation  because  they  are  independent  of  L,  M,  and 
T.     Using  L,  F,  and  T,  the  dimensional  form  of  the  equation  is 


■)' 


which  is  simpler  than  the  first  form.  Indeed  dimensional  equations  based 
on  L,  F,  and  T  are  generally  the  simpler  in  the  case  of  formulas  with  which 
engineers  have  to  deal,  particularly  if  mass  does  not  appear  in  the  formula. 

Showing  that  an  equation  is  homogeneous  does  not  prove  that  it  is  correct, 
but  that  it  may  be  correct;  showing  that  an  equation  is  non-homogeneous 
shows  it  to  be  incorrect.  Since  abstract  numbers  do  not  appear  in  the  dimen- 
sional form  of  an  equation,  the  test  for  homogeneity  does  not  disclose  errors 
in  numerical  coefficients  and  terms,  nor  errors  in  signs. 

(2)  To  express  a  magnitude  in  different  units.  —  Obviously  the  numerical 
value  of  a  given  quantity  changes  inversely  as  the  magnitude  of  the  unit 
used;   thus  a  certain  distance  may  be  expressed  as 

10  mi.,  17,600  yds.,  and  52,800  ft., 

and  plainly  the  numerics  are  respectively  as  i,  1760,  and  5280,  while  the  cor- 
responding units  are  as  5280,  1760,  and  i. 

Let  qi  be  the  known  numerical  value  of  a  quantity  when  expressed  in  the 
unit  ^1,  and  92  the  numeric  (to  be  found)  of  the  same  quantity  expressed  in 
the  unit  Q2;  then 

qi/g2  =  Q2/Q1,     or    gi  =  g-iQi/Qi- 

The  ratio  Q1/Q2  can  be  easily  computed  by  substituting  for  Qi  and  Q2  their 
eguivalents  in  terms  of  fundamental  units;  thus  if  a,  b,  and  c  are  the  dimen- 
sions of  Qi  (and  Q2), 

Qi  =  ki  (Li°Mi*Ti^)     and     Q2  =  ^2(L2''M2''T2=), 

where  Li,  Mi,  and  Ti  are  the  particular  fundamentals  for  Qi;  L2,  M2,  and  T2 
those  for  Q2;  and  ki  and  k2  numerical  coefficients  (very  often  unity).     Finally, 

k2  fuy  /MoV  (I2 

As  an  example,  let  us  determine  how  many  watts  in  10  horse-power.  Since 
Qi  (horse-power)  =  550  ft-lb-sec~S  and  Q2  (watt)  =  10^  ergs  per  sec.  =  10^ 
cm-dyne-sec~^, 

qt;o  ft     lb    sec~^  i;SO/        o\  /         ^^      ^\  /  \        ^ 

gi  =  10^  —  -. zi  =  10^(30.48)  (4.45  X  io5)  (i)  =  7640. 

10' cm  dyne  sec  ^  10^ 

(3)  To  ascertain  the  unit  of  the  result  of  a  numerical  calculation.  —  Substi- 
tute for  the  quantities  the  names  of  the  units  in  which  they  are  expressed, 
and  then  repeat  the  calculation,  treating  the  names  as  though  they  were 


APPENDIX   A  307 

algebraic  quantities.  The  reduced  answer  is  the  name  of  the  unit  of  the 
numerical  answer.  Thus  in  the  formula  for  the  elongation  of  a  rod  due  to  a 
pull  at  each  end,  Pl/AE  (wherein  P  denotes  pull,  I  length  of  the  rod,  A  area 
of  cross-section,  and  E  Young's  modulus  for  the  material),  suppose  that  P  = 
10,000  lbs.,  /  =  50  in.,  .4  =  0.5  in",  E  =  30,000,000  lbs/in^;  the  calculations 
for  elongation  and  name  of  unit  are 

10,000  X  tjo  ,         lbs  X  in  lbs  X  in  X  in^ 

=  0.33,    and     •  0^1.     ,.  2  =      -2  s^  IK.      =  !"• 


0.5  X  30,000,000         ■"^^'  in-  X  lbs/in"  in^  X  lbs 


APPENDIX   B 


MOMENT  OF  INERTIA   AND  RADIUS   OF   GYRATION   OF    PLANE   AREAS* 

§  I.  Elements  of  the  Subject.  —  In  the  subject  of  Strength  of  Ma- 
terials, students  of  engineering  meet  certain  quantities,  formulas  for  which  are 
like 

dA  •  x^, 


P 


where  dA  denotes  elementary  area  and  x  the  distance  of  dA  from  some  line, 
and  the  integration  is  to  be  extended  over  some  finite  area  as  the  cross  section 
of  a  beam,  column,  etc.     In  Chapter  V  of  this  book  there  are  integrals  like 


/ 


dA  'X,     or    dAi- Xi  +  dAi •  X2  +  • 


Since  each  of  the  terms  in  this  summation  is  the  product  of  an  elementary- 
area  and  its  distance  from  some  line  (the  y  coordinate  axis),  each  term  (and 
their  simi)  has  been  called  "moment  of  area;"  this  name  is  in  line  with 
"moment  of  force"  which  is  a  similar  product.     Likewise,  since 


/ 


dA  •  x"^  =  {dAi  •  x{)xi  +  {dA  •  a;2):^2  + 


and  since  each  term  of  the  summation  may  be  regarded  as  the  moment  of  a 
moment,  the  terms  (and  more  particularly  their  sum)  are  called  "second 
moments  of  area."  Thus  these  names  for  the  integrals  are  quite  appropriate. 
But  the  names  are  not  in  general  use;  others  not  so  appropriate  are  more 
common.  The  first  moment  is  generally  called  statical  moment,  and  the 
second  is  generally  called  moment  of  inertia.     This  latter  name  came  into  use 

because  the  integral  named  is  like  a  certain  other  integral,  /  dM  •  r^  (Art.  36) 

which  has  been  previously  called  (with  some  reason)  moment  of  inertia. 
Students  should  recognize  at  once  that  an  area  has  no  inertia,  and  hence 
in  the  ordinary  sense  of  the  words,  no  moment  of  inertia.  There  is  therefore 
no  physical  meaning  whatsoever  attached  to  moment  of  inertia  of  an  area. 
Nevertheless,  the  term  is  so  firmly  established  that  we  will  follow  the  common 
usage.     Thus, 

Moment  of  inertia  of  an  area  with  respect  to  or  about  a  straight  line  (or 
axis)  is  the  sum  of  the  products  obtained  by  multiplying  each  elementary 

*  Writers  on  Strength  of  Materials  usually  refer  to  works  on  Mechanics  for  a  discussion 
of  this  subject,  and  for  that  reason  this  appendix  is  included  herein. 

308 


Appendix  b 


309 


part  of  the  area  by  the  square  of  its  distance  from  the  line.  Engineers  have 
occasion  to  compute  moments  of  inertia  of  plane  areas  only,  and  about  a  line 
which  is  either  in  the  plane  or  perpendicular  to  it.  The  moment  of  inertia 
of  an  area  about  a  Hne  perpendicular  to  the  plane  of  the  area  is  called  a  polar 
moment  of  inertia,  and  the  Une  a  polar  axis. 

The  almost  universal  symbol  for  moment  of  inertia  is  /.  A  subscript  on 
the  symbol  indicates  the  axis  to  which  the  moment  of  inertia  refers;  thus  /;, 
means  moment  of  inertia  about  the  x  axis.  Using  p  to  stand  for  distance  of 
the  elementary  area  dA  from  the  axis  we  have  a  general  formula 

(i) 


/  =    fdA  .  p2  =  J^i .  pi^  +  dA2  •  P2'  + 


In  using  this  formula  care  must  be  taken  to  select  the  elementary  areas  so 
that  all  parts  of  each  are  equally  distant  from  the  axis.  If  this  is  not  done, 
then  the  distance  p  is  uncertain.  This  caution  is  illustrated  in  the  first  exam- 
ple following. 

Each  term  in  the  preceding  series  is  the  product  of  four  lengths;  hence  a 
moment  of  inertia  of  an  area  is  four  "dimensions"  in  length.  The  numerical 
value  of  a  moment  of  inertia  of  an  area  is  usually  computed  with  the  inch  as 
unit  length,  and  the  corresponding  unit  moment  of  inertia  is  called  a  "bi- 
quadratic inch,"  abbreviated  in.* 

Examples.  —  i.  It  is  required  to  ascertain  the  value  of  the  moment  of 
inertia  of  a  rectangle  whose  breadth  and  height  are  b  and  h  respectively,  about 


the  base.  If  we  choose  for  dA  a  strip  parallel  to  the  base  (see  Fig.  i),  then  all 
parts  of  dA  are  equally  distant  from  the  base.  Let  y  =  that  distance;  then 
dA  =  b  dy,  and 


=  J\bdy)f  =  ^^^y''^^  =  ibh'. 


If  we  take  a  vertical  strip  for  elementary  area,  then  dA  =  h  dx.  Now  p,  the 
distance  of  dA  from  the  base,  is  ambiguous;  one  would  naturally  take  it  to  be 
I  h.    Trying  |  /?  we  get 

h=    f\hdx){hhY  =  \h'[x]=lbh\ 
Jo  0 

which  differs  from  the  first  result  and  is  wrong. 


310 


Appendix  b 


2.  It  is  required  to  ascertain  the  value  of  the  moment  of  inertia  of  a  tri- 
angle whose  base  and  height  are  h  and  h  respectively,  about  a  hne  through 
its  centroid  parallel  to  the  base.  For  (any)  elementary  area  we  take  a  strip 
as  shown  in  Fig.  2;  let  y  =  its  distance  from  the  axis,  and  u  =  its  length. 
Then  dA  =  u  dy  =  (b/h)  (f  h  —  y)dy,  and  hence 

/x  =  7  /         ilh-y)  f  dy  =  ^\  bh\ 
lij-\h 

3.  It  is  required  to  ascertain  the  value  of  the  moment  of  inertia  of  a  circle 
whose  radius  is  r  about  a  polar  axis  through  its  center.  Here  it  is  practically 
necessary  to  take  an  elementary  area  of  the  second  order,  as  dx  dy  or  p  dd  •  dp 
in  polar  coordinates.     We  choose  the  latter  (see  Fig.  3).     Then 


Jo  Jo 


(pdd-dp)p'^  =  ^Trr*. 


Radius  of  Gyration.  —  Since  any  moment  of  inertia  of  an  area  is  four  "di- 
mensions" in  length,  it  can  be  expressed  as  the  product  of  the  area  and  a 


Fig. 


length  squared.  It  is  sometimes  convenient  to  so  express  it.  The  length 
is  called  the  radius  of  gyration  of  the  area  about  the  line  to  which  the  moment 
of  inertia  refers;  thus  if  k  and  I  denote  radius  of  gyration  and  moment  of 
inertia  of  an  area  A  about  the  same  Une,  then 

(2) 


k^A 


or 


Vl/A. 


This  length  k  was  called  radius  of  gyration  because  of  the  analogy  between 
it  and  another  length  which  had  been  previously  called  radius  of  gyration. 
This. other  length  is  defined  by  the  equation  k'^AI  =  /,  where  /  is  the  moment 
of  inertia  of  a  body  of  mass  M  about  some  line  (Art.  36).  For  this  length, 
the  term  radius  of  gyration  is  more  or  less  appropriate,  but  for  the  first,  de- 
fined by  equation  (2),  the  term  is  not  appropriate,  except  through  the  analogy. 
It  is  worth  noting  that  the  square  of  the  radius  of  gyration  of  an  area  with 
respect  to  any  line  is  the  mean  of  the  squares  of  the  distances  of  all  the  equal 
elementary  parts  of  the  area  from  that  hne.  For  let  pi,  P2,  etc.,  be  the  dis- 
tances from  the  elements  (dA)  to  the  line,  and  let  n  denote  their  number  (in- 
finite) ;  then  the  mean  of  the  squares  is 

W-\-P2^-\-P3'+    •  •  •)/n=  {pi' dA  +  P2' dA -{- .  .  .  )/ndA  =  I/A  =  k\ 


Appendix  b 


311 


Parallel  Axes  Theorem.  — There  is  a  very  simple  relation  between  the  mo- 
ments of  inertia  (and  the  radiuses  of  gyration)  of  an  area  with  respect  to  par- 
allel axes,  one  of  which  passes  through  the  centroid  of  the  area.  Thus  the 
moment  of  inertia  of  an  area  with  respect  to  any  hne  exceeds  the  moment 
of  inertia  with  respect  to  a  parallel  line  through  the  centroid  by  an  amoimt 
equal  to  the  product  of  the  area  and  the  square  of  the  distance  between  the 
lines.  Or  if  /  and  /  denote  the  moments  of  inertia  respectively,  A  the  area, 
and  d  the  distance  between  the  lines,  then 

I  =  l+Ad\  (3) 

Proofs  of  the  Theorem.  —  (i)  When  the  two  axes  are  in  the  plane  area.  — 
Let  C  (Fig.  4)  be  the  centroid  of  the  area,  U  and  X  the  two  parallel  lines  or 
axes,  and  v  and  y  =  the  ordinates  of  the  elementary  area  dA  from  those  lines 
respectively.     Then 

/  =  ^dA'v"  =JdA(y  +  dy  =JdA  •/  +  2dJdA-y-\-d^JdA. 

Now  I  dA  -y^  =  J;   f  dA  ' y  =  o,  shown  presently;    and  j  dA  =  A;    hence, 

I  =  1  +  Ad^.     dA  •  y  is  the  statical  (or  first)  moment  of  dA  about  CX,  and 

the  sum  of  all  such  terms  a.s  dA  -y  is  the  statical  moment 

of  the   area  about  CX.     Since  this  line  contains  the 

centroid,   the   statical   moment  equals  zero   (Art.    22). 

(2)  When  the  two  lines  are  perpendicular  to  the  area. 

—  Let  0  and  C  (Fig.  5)  be  the  points  where  the  two 

parallel  lines  pierce  the  area,  C  being  the  centroid  of 

the  area.     We  take  OC  for  an  x  coordinate  axis,  and 

the  y  axis  as  shown.     Let  x  and  y  =  the  coordinates 

of  dA ;  then  since  d  =  OC,  the  square  of  the  distance  of 

dA  from  0  is  {x  -\-  df  +  y"^.     Hence  the  moment  of  inertia  of  the  area  with 

respect  to  the  parallel  line  through  0  is 

fdA  [{x  +  df  +  /]  =  JdA  (x2  +  /)  +  2  dJdA  .  X  +  d^JdA. 

Now  x^  +  /  equals  the  square  of  the  distance  from  dA  to  C;  hence 

/  dA{x^  4-  /)  =7;  and  as  already  shown  I  dA  -  x  =  o,  and  j  dA  =  A. 

Therefore  I  =  I  +  A-d^. 

If  we  divide  both  sides  of  equation  (3)  by  A  we  get  I/A  =  I/A  -f  d"^,  or 

k^=~k'i-  d\  (4) 

where  k  and  k  respectively  denote  the  radiuses  of  gyration  of  the  area  with 
respect  to  any  line  and  a  parallel  line  through  the  centroid  and  d  is  the  dis- 
tance between  the  lines.     According  to  this  equation  k  is  always  greater  than 


Fig.  5 


312 


Appendix  b 


d;  that  is,  the  radius  of  gyration  of  an  area  with  respect  to  a  Hne  is  always 
greater  than  the  distance  from  the  line  to  the  centroid  of  the  area.  But  if 
the  line  is  outside  the  area  so  that  d  is  great  compared  with  the  greatest  di- 
mension of  the  area  in  the  direction  of  d^  then  k/d  is  small  compared  to  i 
and  k  equals  d  approximately.  In  such  a  case,  the  moment  of  inertia  equals 
Ad^  approximately. 

The  parallel  axes  theorems  enable  one  to  simplify  many  calculations  on 
moment  of  inertia  and  often  to  avoid  integrations.  Thus  having  found  by 
integration  in  example  i  preceding  (or  otherwise)  that  the  moment  of  inertia 
of  a  rectangle  with  respect  to  its  base  equals  \  h¥  (where  b  and  h  are  base  and 
height  respectively),  we  can  write  at  once  that  the  moment  of  inertia  with 
respect  to  the  median  parallel  to  the  base  is 

\  b¥  -  hh  (i  hy-  =  tV  bh\ 

With  respect  to  a  line  parallel  to  the  base  at  quarter  or  three  quarters  height, 
the  moment  of  inertia  is 


tV 


bh'  +  (bh)  (i  hy  = 


7 
4? 


bh' 


5 


1 1 


In  steel  structural  design  it  is  often  necessary  to  compute  the  moment  of 
inertia  or  radius  of  gyration  of  the  cross  section  of  a  beam  or  column  which 
is  to  be  "built  up"  of  so-called  "structural  shapes,"  about  some  Une  of  the  sec- 
tion.     Fig.   6   represents  the   section  of  a  built-up 
column  consisting  of  a  web  plate  W,  two  side  plates 
S,  and  four  Z  bars.     Manufacturers  of  such  shapes 
publish  "hand  books"  which  include  detailed  infor- 
mation about  the  shape  sections,  —  dimensions,  area, 
position  of  centroid,  moments  of  inertia  and  radiuses 
of  gyration  about  several  lines  through  the  centroid, 
etc.     Thus  for  the  Z  section,  6  X  3I  X  f  inches,  it 
is  given  that  its  area  =  8.63  inches^,  its  moments  of 
inertia  respectively  about  horizontal  and  vertical  axes 
through  its  centroid  (Fig.  6)  =42.12  and  15.44  inches*. 
For  another  example  of  the  use  of  the  parallel  axes  theorem,  we  will  com- 
pute the  moment  of  inertia  of  the  built-up  section  represented  in  Fig.  6  about 


— -^if— 


■7|"- 


-^ 


X 

w 


Fig.  6 


the  X  axis, 
inches)  is 


The  moment  of  inertia  of  the  web-plate  section  (7.75  X  0.75 
tV  7*75  X  0.75'  =  0.27  inches*. 


The  moment  of  inertia  of  the  two  side  plate  sections  (14  X  0.75  inches)  about 
the  X  axis  is 

2  [tV  14  X  0.753  +  (14  X  0.75)  6.752]  =  959.0  inches*, 

6.75  inches  being  the  distance  from  the  centroid  of  either  rectangle  to  the 
X  axis.     The  moment  of  inertia  of  the  four  Z  sections  about  the  x  axis  is 


4  [42.12  -f  (8.63  X  3-375^)]  =  561.7  inches*, 


Appendix  b  3  o 

3.375  inches  being  the  distance  from  the  centroid  of  a  Z  section  to  the  x  axis. 
Hence  the  moment  of  inertia  of  the  entire  section  about  the  x  axis  is 

0.3  +  959.0  +  561.7  =  1 52 1. cinches^. 

While  the  moment  of  inertia  of  a  composite  area  with  respect  to  a  Une  can 
be  fomid  by  adding  the  moments  of  inertia  of  the  component  parts  about  that 
line,  the  radius  of  gyration  of  the  area  cannot  be  found  in  that  way.  To  find 
the  radius  of  gyration  in  such  a  case,  find  the  moment  of  inertia  first,  and  then 
use  k  =  {I /AY'.  For  example,  let  it  be  required  to  find  the  radius  of  gyra- 
tion of  the  cross  section  of  two  6  X  4  X  i  inch  angles,  placed  as  shown  in 
Fig.  7  about  the  line  XX  through  their  centroid.     We  find  in  a  hand  book 


[<._..  4  — >,<- 


m 


i 


Y 

Fig.  7 


6" 


±. 


Fig.  8 


that  the  radius  of  gyration  of  a  single  angle  about  the  line  XX  is  1.85  inches, 
and  that  the  area  of  one  section  is  9  inches 2.  Hence  the  moment  of  inertia 
of  the  pair  about  XX  =  2  (9  X  1.852),  and  the  radius  of  gyration  of  the 
pair  is 


V 


'2  (9  X  1.85-) 
2X9 


=  1.85  inches. 


Three  Rectangular  Axes  Theorem. —The  moment  of  inertia  of  an  area  with 
respect  to  any  polar  axis  (perpendicular  to  the  area)  equals  the  sum  of  the 
moments  of  inertia  of  the  area  with  respect  to  any  two  rectangular  axes  which 
intersect  the  polar  axes  and  he  in  the  area.  If  the  rectangular  axes  and  the 
polar  axis  be  regarded  as  x,  y,  and  z  coordinates  axes  respectively,  then  the 

theorem  can  be  written 

/,  =  /,  +  /,.  (5) 

To  prove  this  theorem  let  x  and  y  =  the  coordinates  of  the  element  dA  (Fig. 
8).     Then  the  distance  of  dA  from  the  z  axis  is  {x^  +  /)^  and  hence 

/,  =  p/l(x2  +  /)  =fdA  '  .v''  +JdA  .  /  =  /^  +  /^. 

If  equation  (5)  be  divided  by  ^,  we  get  at  once 

h^  =  K^  +  ky\  (6) 

where  K,  ky,  and  k^  denote  the  radiuses  of  gyration  with  respect  to  the  x,  y, 
and  2  axes  respectively. 


314 


Appendix  b 


Graphical  Determination  of  the  Moment  of  Inertia  of  a  Plane  Area.  —  If  the 
area  is  quite  irregular  in  shape  so  that  it  cannot  be  divided  into  simple 
parts  whose  moments  of  inertia  are  known,  then  the  method  now  to  be  ex- 
plained may  be  resorted  to  for  finding  the  moment  of  inertia  of  the  irregular 
area  about  any  Hne  in  its  plane.  This  method  is  merely  graphical  integra- 
tion.    Let  the  area  at  the  left  in  Fig.  9  be  the  irregular  area  and  XX  the  line 


1 
1 

k- 

lY 

2" 
1 

\ 

1 

A 

io 

■'-             \  X 

Fig.  10 


about  which  the  moment  of  inertia  of  the  area  is  required, 
width  of  the  area,  parallel  to  XX,  at  any  point  of  the  figure; 
distance  of  the  point  or  width  from  XX.     Then 


Let  w 
and  y 


the 
the 


/  (w  dy)  y^  =  j  (wy^)  dy  =  \  w'  dy, 


where  w'  is  merely  an  abbreviation  for  wy-.  Now  suppose  we  multiply  several 
widths  w  by  the  square  of  the  corresponding  distances  y,  lay  off  the  products 
wy^  to  any  convenient  scale  from  a  perpendicular  to  XX  as  shown,  and  then 
draw  a  smooth  curve  through  the  ends  of  the  ordinates  or  distances  wy^  or  w'. 
The  area  between  this  smooth  curve  and  the  perpendicular  equals 


/ 


w'  dy,  and  hence  it  represents  /. 


Evidently,  the  modified  area,  as  we  may  call  it,  must  be  interpreted  according 
to  some  scale  as  we  will  explain  in  connection  with 


w 

y 

wy- 

w 

y 

wy^ 

2.00 

0 

0 

1-54 

1-25 

2.40 

1 .96 

0.25 

0.  122 

1.30 

I  5° 

2.92 

1. 91 

0.50 

0.477 

0.93 

I -75 

2.8s 

1.83 

0.75 

1.03 

0 

2.00 

0 

1. 71 

1 .00 

I. 71 

A  numerical  example.  —  Instead  of  an  irregular  figure,  we  take  a  regular  one 
so  that  we  can  compute  its  moment  of  inertia  by  an  exact  method  also,  and 
thus  check  the  graphical  method.  We  will  compute  the  moment  of  inertia 
of  a  circular  quadrant  (Fig.  10)  of  2-inch  radius  about  one  of  the  straight  sides. 
We  have  taken  nine  widths  w,  see  adjoining  table.  At  the  right-hand  of  the 
figure  we  have  laid  off  the  products  wy^,  or  w',  to  the  scale  i  inch  =  5  inches^. 


Appendix  b  315 

The  new  area  is  0.25  inches^.  Since  the  scale  of  the  quadrant  is  i  inch  = 
2.5  inches,  the  scale  of  the  new  area  is  i  square  inch  =  5  X  2.5  =  12.5  inches*. 
Hence,  the  construction  gives  0.25  X  12.5  =  3.12  inches'*  as  the  moment 
of  inertia  desired.     The  exact  formula   i^^Trr*)  gives  3.142   inches*. 

§  2.  Formulas  for  Moment  of  Inertia  and  Radius  of  Gyration  for 
SOME  Special  Cases.  —  In  the  following,  /  and  k  are  symbols  for  moment 
of  inertia  and  radius  of  gyration  respectively.  Only  a  few  formulas  for  k  are 
stated;  in  any  case  k  can  be  computed  from  V//area. 

Rectangle.  —  Let  b  =  base  and  h  =  altitude.  About  a  line  through  the 
center  parallel  to  b,  /  =  tV  b¥-  About  a  line  through  the  center  parallel 
to  h,  I  =  tV  hb\  About  the  base  b,  I  =  I  b¥.  About  the  side  h,  I  =  \  hb\ 
About  a  diagonal,  1  =  1  bW/Qf-  +  h"^)-  About  a  line  through  the  center 
perpendicular  to  the  rectangle,  /  =  yV  (^^^  +  ^^^)- 

Square.  —  Make  b  =  h  m  foregoing.  The  moment  of  inertia  for  all  axes 
in  the  plane  of  the  square  and  passing  through  the  center  is  ^^  h^,  where  h 
is  the  length  of  one  side  of  the  square. 

Hollow  Rectangle.  —  Let  B  and  b  =  outer  and  inner  breadths,  and  H  and 
h  =  outer  and  inner  heights.  About  an  axis  parallel  to  B  and  b  and  passing 
through  the  center,  I  =  tV  (BH^  -  bh^). 

Triangle.  —  Let  b  =  base  and  h  =  altitude.  About  the  base,  /  =  ^\  bh^. 
About  a  line  through  the  centroid,  and  parallel  to  the  base,  /  =  5^  bh^.  About 
a  line  through  the  vertex  and  parallel  to  the  base,  1  =  1  bh^. 

Regular  Polygon.  —  Let  A  =  area,  R  =  radius  of  circumscribed  circle, 
r  =  radius  of  inscribed  circle,  and  s  =  length  of  a  side.  About  any  axis 
through  the  center  and  in  the  plane  of  the  polygon,  /  =  0^4  A  (6  R^  —  s"')  = 
5^5  y4(i2  r^  +  5^).  About  a  line  perpendicular  to  the  plane  of  the  polygon 
passing  through  the  center,  /  =  double  the  preceding  /. 

Trapezoid.  —  Let  B  =  long  base,  b  =  short  base,  and  h  =  altitude.  About 
the  long  base,  /  =  yV  (^  +  3  W^^-  About  the  short  base,  /  =  tV  (3  ^  +  ^)^^- 
About  a  line  through  the  centroid  and  parallel  to  the  bases. 

Circle.  —  Let  d  =  diameter  and  r  =  radius.  About  a  diameter,  I  = 
^^Tj-d*  =  lirr*;  k  =  \d  =  \r.  About  a  line  through  the  center  and  per- 
pendicular to  the  circle,  /  =  ^^i^d'^  =  ^tt/-*;    ^  =  V|  J  =  Vf  r. 

Semicircle.  —  Let  d  =  diameter  and  r  =  radius.  About  the  bounding 
diameter  or  about  the  line  of  symmetry,  /  =  jis  7r<i*  =  |  irr*.  About  a  line 
through  the  centroid  and  parallel  to  the  bounding  diameter, 

7  =  (qtt^  —  64)  JVii52  7r  =  0.00686  c/*  =  0.1 10  r*. 

Hollow  Circle.  —  Let  D  and  d  =  outer  and  inner  diameters,  and  R  and 
r  =  outer  and  inner  radiuses.  About  a  diameter,  I  =  ^^^  t  (D*  —  d*)  = 
i(/?4_^).     ^  ^  1  (£)2  ^  j2)5  =  1  (7^2  _|_  y2)j_     About    a    line    through    the 

center  and  normal    to  the  circle,   I  =  ^tv{D^  -  d*)  =  ^  ir  {R^  -  r*) ;   k  = 


oj5  Appendix  b 

Circular  Segment.  —  Let  A  =  area  of  the  segment,  r  =  radius  of  the  arc, 
and  2  a  =  the  angle  subtended  at  the  center  by  the  arc.  About  the  line  of 
symmetry  of  the  segment, 

/  =  J  Ar^  [i  —  3  (sin^ «  cos  a)/{a  —  sin  a  cos  a)]. 
About  the  diameter  of  the  circle  which  is  parallel  to  the  straight  side  of  the 

segment, 

I  —  \  Ar'^\\  +  f  (2  sin^acosQ;)/(a  —  sin  a  cos  a)]. 

Circular  Sector.  —  Let  A  =  area  of  the  sector,  r  =  radius  of  the  arc,  and 
2  a  =  the  angle  subtended  at  the  center  by  the  arc.  About  the  line  of  sym- 
metry of  the  sector, 

I  —  i  Ar~  (i  —  sin  a  cos  a/a). 

About  a  line  through  the  center  perpendicular  to  the  Une  of  symmetry  and 
in  the  plane  of  the  sector, 

I  —  \  Ar-  {1  -{■  sin  a  cos  a/ a). 

About  a  line  through  the  center  of  the  arc  and  perpendicular  to  the  plane  of 
the  sector,  I  =  \  Ar"^. 

Parabolic  Segment  bounded  by  an  arc  of  a  parabola  and  a  chord  which  is 
perpendicular  to  the  axis  of  the  parabola.  Let  a  =  distance  from  the  vertex 
to  the  chord  and  b  =  length  of  the  chord.     About  the  axis  of  the  parabola 

1  =  3V  aP.     About  the  tangent  at  the  vertex  of  the  parabola,  /  =  f  ba^. 
Ellipse.  —  Let  2  a  and  2  b  =  lengths  of  the  axes  of  the  ellipse.     About  the 

2  a  axis,  /  =  i  Trab^.     About  the  2  b  axis,  /  =  i  Tba^.     About  a  line  through 
the  center  and  perpendicular  to  the  ellipse,  /  =  5  -Kabiar  +  b'). 

§  3.  Product  of  Inertia  and  Principal  Axes.  —  Preparatory  to  another 
matter,  we  will  now  discuss  briefly  a  quantity  called  product  of  inertia  of  a 
plane  area  with  respect  to  two  rectangular  coordinate  axes  in  the  plane.  By 
this  term  is  meant  the  sum  of  all  the  products  obtained  by  multiplying  each 
elementary  area  by  its  coordinates.  Thus  if  dAi,  dAi,  etc.,  denote  (second 
order)  elements  of  the  area,  and  (xiji),  (xaVa),  etc.,  denote  their  coordinates 
respectively,  then  the  product  of  inertia  is  dAiXiji  +  dA20C2y2  +    .  .  .  ,  or 


J. 


=    /  dAxy,  (i) 


J XV  being  the  symbol  which  we  shall  use  for  product  of  inertia  with  respect  to 
axes  X  and  y.  It  is  plain  from  the  definition  and  expression  that  a  unit  prod- 
uct of  inertia  is  four  "dimensions"  in  length.  Like  moments  of  inertia  we 
will  express  products  of  inertia  in  biquadratic  inches. 

Unlike  a  moment  of  inertia,  a  product  of  inertia  may  be  zero  or  negative. 
For  example,  the  product  of  inertia  of  the  rectangle  (Fig.  11)  with  respect  to 
the  axes  OX  and  OY  is  zero,  which  may  be  shown  as  follows:  for  every  elemen- 
tary area  whose  coordinates  are  (c,  6),  there  is  one  whose  coordinates  are 
(a,  —  b),  and  hence  the  product  of  inertia  of  the  pair  is  dA  '  ab  —  dA  •  ah  =  o\ 
therefore  the  product  of  inertia  of  the  entire  area  is  zero. 


Appendix  b 


317 


The  product  of  inertia  of  the  rectangle  with  respect  to  the  axes  O'X'  and 
O'Y'  is  negative;  for  one  of  the  coordinates  of  each  element  is  negative  and 
the  other  is  positive,  and  hence  the  product  of  inertia  of  each  element  is  nega- 
tive. Even  for  different  pairs  of  axes  with  the  same  origin  the  product  of 
inertia  of  an  area  may  be  positive,  zero,  or  negative.  Thus,  the  product  of 
inertia  of  the  triangle  (Fig.  11)  for  the  axes 
shown  is  zero.  If  the  axes  be  turned  clock- 
wise about  0  slightly,  then  the  product  of 
inertia  is  negative;  and  if  turned  counter 
clockwise  slightly,  then  it  is  positive.  As  will 
be  shown  presently  there  is  always  one  pair 
of  axes  through  each  point  of  an  area  with 
respect  to  which  the  product  of  inertia  is  zero, 
and  this  pair  is  of  prime  importance  in  certain 

particulars.  If  the  area  has  a  line  of  symmetry,  then  some  of  the  pairs  of 
axes  for  which  the  product  of  inertia  of  the  area  is  zero  can  be  identified 
easily;  indeed  for  such  an  area,  the  product  of  inertia  is  zero  with  respect 
to  the  axis  of  symmetry  and  any  line  (in  the  area)  perpendicular  to  that 
axis.  For  if  we  think  of  the  elementary  areas  grouped  into  pairs  sym- 
metrical with  respect  to  the  axis  of  symmetry,  then  we  see  that  the  product 
of  inertia  of  each  pair  —  and  hence  that  of  the  entire  area  —  equals  zero. 

Parallel  Axes  Theorem  for  Products  of  Inertia.  —  There  is  a  simple  relation 
between  the  products  of  inertia  of  an  area  with  respect  to  two  parallel  sets 
of  coordinate  axes,  the  origin  of  one  set  being  at  the  centroid  of  the  area.  It 
is  expressed  by 

/  =  /  -f  Axy,  (2) 

where  /  =  the  product  of  inertia  about  the  axes  through  the  centroid,  /  =  the 

product  about  the  other  pair,  A  —  the  area,  and  (xy)  = 
the  coordinates  of  the  centroid  with  respect  to  the  second 
set  of  axes.  To  deduce  equation  (2),  let  C  (Fig.  12)  be 
the  centroid  of  the  area,  0  any  other  point,  CU  and  CV 
one  set  of  axes,  OX  and  OY  another  parallel  set,  {u,  v) 
and  {x,  y)  =  the  coordinates  of  any  elementary  area  dA 
with  respect  to  these  sets  of  axes  respectively.  Then 
X  =  u  -\-  X  and  y  =  v  -^  y;  also 


Fig.  12 


J  =  I  dA(u  +  x){v  -{-  y)  =^    I  dA  '  uv  -\-  xy  j  dA  -\-  x  j  vdA  -\-  y  I  udA. 
ow  /  dA  'UV  =  J;  xy  j  dA  =  Axy;  j  v  dA  and  I  ; 


u  dA  —  the    statical    mo- 


ments of  A  about  CU  and  CV  respectively,  and  these  moments  equal  zero 
since  these  lines  contain  the  centroid  (Art.  22).     Therefore  J  =  J  -\-  Axy. 
We  will  now  illustrate  by  determining  the  product  of  inertia  of  the  angle 


•c. 


-2 1 3  Appendix  b 

section  (7  X  32  X  i)  shown  in  Fig.  13  with  respect  to  the  axes  CX  and  CY. 
Imagine  the  section  divided  into  two  rectangles  as  shown;  their  areas  are 
3.5  X  I  =  3-5  inches^,  and  6X1=6  inches^.  The  coordinates  of  the  cen- 
troids  of  these  areas  with  respect  to  the  axes  CX  and  CY  are  (0.79,  —  2.21) 
.y  and  (—  0.46,  1.29)  respectively.     Now  the  product  of  inertia 

.•0.96  of  each  rectangle  with  respect  to  axes  through  its  centroid 

parallel  to  CX  and  CF  is  zero;   therefore  according  to  the 

-  parallel  axes  theorem,  the  product  of  inertia  of  the  entire 
27/    section  about  CX  and  CF  is 

k-  .  jr- .  /""  [o  +  3-5  (0-79)  (-  2.21)]  +  [o  +  6  (-  0.46)  (1.29)] 

Fig.  13  =  -  6.1 1  -  3.56  =  -  9.67  inches*. 

Inclined  Axis  Theorem  for  Moment  of  Inertia. —  Let  OX  and  OY  (Fig.  14) 
be  any  two  rectangular  axes  in  the  area  and  OU  and  OV  another  pair,  XOU 
being  any  angle  a.     It  is  plain  from  the  figure  that 

1}  =  y  cos  a  —  X  sin  a,     and     u  =  ysina  -{-  x  cos  a. 

If  these  values  for  u  and  v  be  substituted  in  /„  =  /  dA  •  v^,  it  will  be  found  on 

simplifying  that 

lu  =  Ix  cos^  a  +  /y  sin-  a  —  /ly  sin  2  «  .  .   .  (3) 

With  this  equation  it  is  possible  to  find  the  moment  of  inertia  of  an  area  with 
respect  to  an  axis  through  any  point  in  the  plane,  if  the  moments  and  the 
product  of  inertia  of  the  area  with  respect  to  two  rectangular  axes  through 
the  point  are  known. 

Obviously,  the  moment  of  inertia  of  an  area  with  respect  to  different  Unes 
through  the  same  point  are  unequal  in  general.  We  will  show  presently  that, 
generally,  there  is  one  Hne  for  which  the  moment  of  inertia  is  greater,  and  a 
line  for  which  the  moment  of  inertia  is  smaller  than  for  any  other  Une  through 
the  point;  also  that  these  two  lines  are  at  right  angles  to  each  other.  They 
are  called  the  principal  axes  of  the  area  for  the  particular  point;  and  corre- 
sponding to  those  axes  we  speak  of  the  principal  moments  of  inertia  and  principal 
radiuses  of  gyration  of  the  area  for  the  point.  The  condition  for  a  maximum  or 
minimum  value  of  /„  is  that  dljda  =  o.     Now  from  (3), 

dl/da  =  —  2  /x  sin  a  cos  a  +  2  ly  sin  a:  cos  a  —  2  J^y  cos  2  a. 
Let  us  denote  by  a  the  value  of  a  which  makes  this  zero.     Then  we  have 

—  Ix  sin  2  a  -{■  ly  sin  2  a'  —  2  Jxy  cos  2  a'  =  o,  or 

tan  2a'  =  J   J^^j  '  (4) 

ly  ix 

In  general,  this  equation  gives  two  values  of  2  a    differing  by  180  degrees; 
hence  two  values  of  a  differing  by  90  degrees.     These  two  values  of  a  fix  two 


Appendix  b 


319 


lines  (w  axes)  which  are  the  principal  axes  for  the  point  under  consideration. 
If  Jxy  =  o  and  ly  =  h,  then  equation  (4)  is  ambiguous.  For  such  case,  equa- 
tion (3)  shows  that  lu  =  h  =  ly',  that  is,  /„  does  not  depend  on  the  inclina- 
tion, and  there  is  no  maximum  or  minimum  value  for  any  axis  through  the 
point. 

The  condition  expressed  by  equation  (4),  for  locating  the  principal  axes, 
can  be  stated  somewhat  differently.     Referring  to  Fig.  14,  it  will  be  seen  that 


Juv  =    I  dA'UV  =  ^  {Ix  —  ly)  sin  2  Q!  +  Jxy  COS  2  a. 


(5) 


Apparently  this  may  equal  zero  for  certain  values  of  a;  indeed  if  we  equate 
it  to  zero  we  will  arrive  at  equation  (4).  Hence,  the  principal  axes  are  such  a 
pair  for  which  the  product  of  inertia  is  zero. 


Fig.  14 


Fig.  is 


For  an  example,  we  will  locate  the  principal  axes  of  the  section  shown  in 
Fig.  13,  for  the  point  C,  it  being  given  that  Ix  =  45-37  and  ly  =  7.53  inches*. 
In  the  preceding  example,  it  is  shown  that  Jxy  =  —  9.67  inches*.  Therefore 
according  to  equation  (4), 

tan  2  a   =  2  (-  9-67)/(7-53  -  45-32)  =  0.5118;  hence 
2  a' =  27°  6' or  207°  6',     and     a' =  13°  33' or  103°  33'. 


Substituting  these  two  values  successively  in  equation  (3),  we  find  as  the  prin- 
cipal moments  of  inertia 

71  =  47.70,     and    /2  =  5.20  inches'*. 

There  is  a  simple  graphical  construction  for  the  radius  of  gyration  of  an 
area  about  any  line  through  a  given  point,  if  the  principal  axes  and  radiuses 
of  gyration  of  the  area  for  that  point  are  known.  Let  O  (Fig.  15)  be  the  point 
(area  not  shown),  OP  the  line,  OX  and  OY  the  principal  axes,  and  kx  and  ky 
the  principal  radiuses  of  gyration  respectively.  We  draw  two  circles  with 
centers  at  0  and  radiuses  equal  to  kx  and  ky-,  and  we  call  the  intersection  of 
these  circles  with  OP,  A  and  B  respectively.  We  draw  Hnes  through  A  and  B 
parallel  to  OY  and  OX  respectively  and  call  their  intersection  C.  Then  OC 
equals  the  desired  radius  of  gyration  (about  OP).  For  when  the  axes  x  and  y 
of  equation  (3)  are  principal  axes,  Jxy  =  o  and 


Ix  cos^  a  -\-  ly  sin-  a, 


or     ku^  = 


kx^  cos'*  a  -{■  ky^  sin  2  a; 


320 


APPEhfDIX  B 


but  OD  (Fig.  15)  ^  kx  cos  a  and  CD  =  kySma,  and  hence  {ODf  +  (CD)^, 
which  equals  {OCf,  =  k^^  cos^ «  +  ky^  sin^  a  =  ku\  or  OC  =  ^u- 

§  4.  Inertia  Curves.  —  By  means  of  a  certain  {inertia)  circle,  we  can 
locate  principal  axes,  find  principal  moments  of  inertia,  etc.,  —  in  short,  do 
graphically  what  we  did  algebraically  in  the  preceding  section.  We  will  now 
show  how  to  draw  and  use  this  circle;  proof  of  the  method  is  supplied  later. 

Let  the  shaded  portion  of  Fig.  16  be  the  area  under  consideration.     To 


Fig.  16 


Fig.  17 


draw  the  required  circle  we  must  know  (as  in  §  3  to  apply  equations  3  and  4) 
the  moments  of  inertia  of  the  area  about  two  rectangular  axes  through  the 
point  under  consideration  and  the  product  of  inertia  about  those  axes;  we 
will  suppose  these  quantities  {h,  ly,  and  J^cy)  to  have  been  determined.  First 
we  lay  off  OA  and  OB  to  represent  h  and  /„  respectively,  according  to  some 
convenient  scale;  draw  BC  from  B  parallel  to  OF,  and  make  BC  =  Jxy  (re- 
quirmg  that  BC  be  drawn  in  the  positive  or  negative  y  direction  according  as 
7x2/  is  positive  or  negative);  we  bisect  AB  in  Q,  and  then  draw  the  circle  with 
center  at  Q  and  radius  equal  to  QC.  This  is  the  inertia  circle  of  the  area  for 
the  axes  OX  and  OY.  If  we  letter  the  intersections  of  the  circle  with  OX  say 
M  and  N,  then  the  principal  axes  for  0  are  parallel  to  CM  and  CN,  and  the 
corresponding  (principal)  moments  of  inertia  are  equal  to  OM  and  ON,  ac- 
cording to  the  scale  used.  To  find  the  moment  of  inertia  of  the  area  about 
any  line  through  0  as  OC/:  — draw  a  secant  through  C  parallel  to  OU,  and 
mark  its  intersection  with  the  circle  D;  from  D  draw  a  line  parallel  to  the  y 
axis  and  mark  its  intersection  with  the  x  axis  E;  then  OE  equals  the  desired 
moment  of  inertia  /„.  Incidentally  we  may  note  that  ED  represents  /„„,  the 
product  of  inertia  of  the  area  with  respect  to  OU  and  OV. 

We  will  prove  first  that  the  construction  for  /„  is  correct.     Equation  (3) 
can  be  written 

lu  =  {h  cos  a  —  Jxy  sin  a)  cos  a  +  (ly  sin  a  —  Jxy  cos  a)  sin  a, 

and  this  form  suggests  the  proof.     Since  (Fig.  17)  h  =  OA,  ly  =  OB,  and 
J^  =  BC  =  AC, 

lu  =  (OA  cos  a  —  AC  sin  a)  cos  a  +  (OB  sin  a  -  BC  cos  a)  sin  a 
=  (Oa  —  ac')  cos  a  +  (Bb  —  Be)  sin  a 
=  Oc'  cos  a+{bc^  c'D)  sin  a  =  Oe-\-{dD  =  eE)  =  OE. 


Appendix  b 


321 


If  we  imagine  OU  to  turn  about  O  clockwise,  say,  CD  (drawn  parallel  to 
OU)  turns  about  C;  and  OE  (and  therefore  /„)  increases.  The  greatest 
value  of  lu  (larger  principal  moment  of  inertia)  obtains 
when  £  is  at  M;  then  D  is  at  M,  and  the  corresponding 
principal  axis  is  parallel  to  CM.  If  we  imagine  0^  to 
turn  counter  clockwise,  OD  turns  and  OE  (or  /„)  gets 
smaller.  The  smallest  value  of  /„  (the  lesser  principal 
moment  of  inertia)  obtains  when  £  is  at  iV;  then  D  is  at 
N  and  the  corresponding  principal  axis  is  parallel  to  CN. 

Inertia  Ellipse.  —  Let  OX  and  OY  (Fig.  18)  be  the 
principal  axes  of  an  area  (not  shown)  for  the  point  0,  ki 
and  ^2  respectively  =  the  radiuses  of  gyration  of  the  area 
with  respect  to  those  axes,  and  OA  =  ^2  and  OB  =  ki, 
then  the  ellipse  AB  is  called  the  inertia  ellipse  of  the  area  for  the  point  O. 
Let  r  =  any  radius  as  OP,  k  =  the  radius  of  gyration  of  the  area  about  OP, 
and  p  —  the  perpendicular  from  0  to  either  tangent  parallel  to  OP;  then, 
as  will  be  shown  presently, 

k  =  kiki/r  =  p.  (i) 

Since  the  coordinates  of  P  are  r  cos  a  and  r  sin  a,  the  equation  of  the  ellipse 
can  be  written 


Fig.  18 


r"^  cos^  a 


ko^ 


+ 


r^  sin^  a 


h' 


=  I,     or 


r^  = 


h^h^ 


k^  cos^  a  +  k'^  sin^  a 


It  follows  from  equation  (3),  §  3,  that  k^  =  k^  cos^  a  +  ^2^  sin^  a;  hence  r^  = 
kik^/k"^,  ox  k  =  kiki/r.  One  of  the  well-known  properties  of  the  ellipse  is 
that  the  product  of  any  radius  and  the  perpendicular  from  the  center  to 
either  tangent  parallel  to  that  radius  is  constant;  that  is  rp  =  kiki]  hence 
kiki/r  =  p. 


PROBLEMS;   SET   A 


The  number  in  Parentheses  following  a  problem  number  refers  to  the  article 
which  pertains  to  that  problem. 

i-(3).  Compound  the  80  and  no  lb.  forces  (Fig.  i)  by  means  of  the  parallelogram 
law.  (To  describe  the  line  of  action  of  the  resultant,  note  where  it  cuts  edges  of  the 
square  board.     Use  scales  of  about  4  ins.  and  40  lbs.  to  the  inch.)* 

2-(3).  Compound  the  50  and  60  lb.  forces 
(Fig.  i)  by  means  of  the  triangle  law.  (Make 
the  vector  diagram  separate  from  the  space 
diagram,  and  use  standard  notation.) 

3-(3).  Compound  the  60  and  70  lb.  forces 
(Fig.  i)  algebraically.  (Specify  the  direction 
of  the  resultant  by  means  of  the  angles  between 
it  and  the  two  given  forces.) 

4-(3).    Compound  the  50  and  90  lb.  forces 

(Fig.  i). 

5-(3).  Resolve  the  40  lb.  force  (Fig.  i)  into 
two  components,  one  parallel  to  the  70  lb.  force 
and  one  vertical,  by  a  graphical  method. f 

6-(3).   Resolve  the  100  lb.  force  (Fig.  i)  into  two  components,  one  of  which  acts 
in  the  lower  edge  of  the  square  and  the  other  through  the  upper  right-hand  corner. 
7^(3).   Resolve  the  no  lb.  force  (Fig.  i)  into  horizontal  and  vertical  components. 
8-(3).   Draw  a  square  and  number  the  corners   i,  2,  3,  and  4  consecutively. 

Imagine  a  force  of  100  lbs.  to  act  in  I2  and  in  the 
direction  '12'.  Resolve  it  into  components  acting  in 
the  other  three  sides. 

9-(4).  Compound  the  40,  50,  60,  and  70  lb. 
forces  (Fig.  i)  graphically.  (Do  not  draw  the  force 
polygon  in  the  space  diagram;  use  standard  nota- 
tion.) 

io-(4).  Compound  the  70,  90,  100,  and  no  lb. 
forces  (Fig.  i)  algebraically.  (Specify  the  direction 
of  the  resultant  by  means  of  the  angle  between  it 
and  the  horizontal.) 

ii-(4).    Compound  the  four  forces  of  the  cube  in 
Fig.  2. 
i2-(s).    Compute  the  moment  of  the  60  lb.  force  (Fig.  i)  about  point  i. 

*  Complete  composition  requires  that  the  magnitude,  line  of  action,  and  sense  of  the 
resultant  be  determined. 

t  Complete  resolution  requires  that   the  magnitude,  line  of   action,  and  sense  of  each 
component  be  determined. 

32J 


VT 


/ 

j/sOlbs. 

30lb5'^'' 

-7 

/ 

/ 

""/ 

/ 

• 

v 

K- 
l 


J'- 


—  >• 


Fig.  2 


324 


i3~(5)-  Compute  the  moment  of  the  40  lb.  force  (Fig.  i)  about  point  2,  making  use 
of  the  principle  of  moments. 

14.  A  certain  chimney  is  150  ft.  high  and  weighs  137,500  lbs.  Suppose  that  it  is 
subjected  to  a  horizontal  wind  pressure  of  54,000  lbs.,  uniformly  distributed  along 
its  height.  Determine  where  the  hne  of  action  of  the  resultant  of  the  weight  and 
pressure  cuts  the  ground. 

15.  Fig.  3  represents  the  cross  section  of  a  masonry  dam.  It  weighs  150  lbs/ft' 
and  the  water  pressure  against  it  is  112,500  lbs.  per  foot  length  of 
dam.  The  resultant  pressure  acts  at  right  angles  to  the  face  of 
the  dam  and  20  ft.  above  its  base.  The  center  of  gravity  of  the 
cross  section  is  11.46  ft.  from  the  face  of  the  dam  and  24  ft.  above 
the  base.  Find  where  the  resultant  of  the  weight  and  the 
pressure  cuts  the  base. 

i6-(5).  Imagine  a  clockwise  couple  of  2  ft-lbs.  to  act  on  the 
square  board  of  Fig.  i.  Then  compound  the  couple  and  the 
40  lb.  force. 

i7-(5).   Fig.  4  represents  a  3  ft.  pulley  on  the  end  of  a  shaft; 
the  pulley  is  subjected  to  a  pull  of  100  lbs.  applied  tangentially 
as  shown. 
Resolve  the  force  into  a  force  acting  through  the  center  of  the  pulley 
and  a  couple. 

i8-(6).  Compound  the  four  forces  (wind  pressures)  represented  in 
Fig.  5.  (Be  prepared  to  give  the  incUnation  of  the  resultant  and  the 
point  where  the  line  of  action  cuts  the  floor.) 

i9-(6).  Fig.  6  represents  one-half  of  an  arch  and  certain  loads 
applied  to  it.     Pi  =  4000,  P2  =  5000,  P3  =  6000,   and   P4  =  10,000 


/S,000 


53,000lb5.    SZOOOIbs. 


I8,500,^^_  I8,500i^^_ 


.^Sl 


37,000/bs. 


< gQ' - •>1  (<--7-->|<-5^>K 76-  ■>f J/  8 •*! 

Fig.  5  Fig.  7 

lbs.;  their  inclinations  are  0°,  3°,  8°,  and  12°  respectively;  the  coordinates  of  points 
I,  2,  3,  and  4  are  (1.6,  o.i),  (4.9,  0.7),  (8.4,  2.1),  and  (12.8,  4.8),  all  in  feet.  Com- 
pound the  four  load  by  the  second  method.  (Specify  the  line  of  action  of  the  result- 
ant by  means  of  the  angle  between  it  and  the  x  axis  and  the  intercept  on  that  axis.) 

20-(7).   Determine  the  resultant  of  the  locomotive  wheel-loads  (Fig.  7). 

2i-(7).   Determine  the  resultant  of  the  loads  described  in  Prob.  19  algebraically. 

22-(8).    Compute  the  moments  of  each  of  the  forces  represented  in  Fig.  2  about 
the  X,  y,  and  z  axes. 


325 

23-(8).  Determine  the  resultant  of  the  three  couples  acting  on  the  4  ft.  cube  repre- 
sented in  Fig.  8.  (Specify  the  plane  of  the  resultant  by  means  of  the  angles  which 
a  normal  to  the  plane  makes  with  the  edges  of  the  cube.^) 

24-(9).    Determine  the  resultant  of  all  except  the  300  lb.  forces  (Fig.  8). 

25-(io).    State  what  you  can  about  the  resultant  in  the  following  cases: 

(a)  A  system  of  coplanar  concurrent  forces  for  which  SF„  =  0;  for  which 

SMa  =  0;  for  which  the  force  polygon  closes. 

(b)  A  system  of  noncoplanar  concurrent  forces  for  which  SF^  =  o. 

(c)  A  system  of  coplanar  parallel  forces  for  which  SF^,  =  o;   for  which 

zMa  =  o;  for  which  the  force  polygon  closes. 

(d)  A  system  of   coplanar  nonconcurrent  nonparallel  forces  for  which 

ZF^  =0;  for  which  zFx  =  2F„  =  0;  for  which  SM„  =  sMb  =  o. 


Fig.  8 


Fig.  9 


Fig.  10 


26-(ii).  A  and  B  (Fig.  9)  are  two  smooth  cylinders  supported  by  two  planes  as 
shown.  A  weighs  200  lbs.  and  B  100  lbs.;  the  diameter  of  ^  is  6  ft.  and  of  B  10  ft.; 
a  =  30°.     Determine  the  pressures  on  the  planes  and  that  between  the  cylinders. 

27-(ii).  Fig.  10  represents  two  wedges;  a  =  70°  and  0  =  40°.  A  push  P  of 
1000  lbs.  can  sustain  what  load  Q  if  all  rubbing  surfaces  are  smooth? 

28-(ii).  The  chains  AB  and  AC  (Fig.  11)  are  5  ft.  long.  When  BC  =  8  ft.  and 
the  suspended  load  W  =  2  tons,  what  is  the  tension  on  each  chain?  If  the  safe  pull 
for  each  chain  is  3  tons,  how  large  may  the  spread  BC  be? 


Fig.  II 


29-(ii).  Two  bars  AB  and  CD  (Fig.  12)  are  connected  by  a  pin  at  A  and  to  a 
floor  by  pins  B  and  C.  BC  =  8  ft.,  AB  =  AC  =  $  ft.,  and  AD  =  8  ft.  A  weight 
of  100  lbs.  is  suspended  from  D.     Determine  the  pin  pressures  at  yl,  B,  and  C. 

30-(ii).  A  carrier  is  arranged  as  shown  in  Fig.  13.  The  bar  AB  connecting  the 
axles  of  the  wheels  is  24  ins.  long.    The  bars  AC  and  CB  are  each  30  ins.  long     There 


326 

is  a  load  of  1200  lbs.  at  C. 

g///////////////////////////////////////////// 

C 

E 
A" 

D 

r 

^ 

^A B4 

Fig.  14 


be  stalled  in  the  position  shown,  ^  =  60 
degrees,  and  the  steam  pressure  150  lbs 
/in^.  Determine  the  push  on  the  connect- 
ing rod  BC  and  the  pressure  against  the 
cross-head  guide  D. 
33-(ii).  The  beU-crank  ^BC  (Fig.  16) 


Determine  the  compression  in  AB  and  the  tension  in  ^C 
and  BC. 

3i-(ii).  AB  (Fig.  14)  is  a  bar  suspended  from  a  ceiling 
by  means  of  vertical  ropes  AC  and  BD.  The  middle 
points E  and  F  are  connected  by  another  rope.  AB  =  AC 
=  BD  =  8  f t.  A  vertical  force  P  is  applied  at  the  middle 
G,  deflects  the  ropes  as  shown  by  the  dotted  hues,  and 
raises  the  bar.  How  large  must  P  be  to  support  the  bar 
(weighing  1000  lbs.)  6  ins.  above  its  original  position? 

32-(ii).  The  cylinder  of  the  steam  engine  (Fig.  15)  is 
10  ins.  in  diameter,  the  crank  AB  \s  $  ins.  long,  and  the 
connecting  rod  5C  is  15  ins.  long.     Assume  the  engine  to 

-\ 

v~~M& 

^1        '^^  \        A 


/ 


Fig.  is 


Fig.  16 


is  pinned  to  a  wall  at  ^ ;  a  cyUnder  G  is  suspended  by  means 
of  a  cord  from  D  as  shown;  BD  =  4  ins.  The  cylinder  weighs 
80  lbs.  and  is  smooth.  Determine  all  the  forces  which  act  upon 
the  bell-crank. 

34-(ii).  Fig.  17  represents  a  riveting  machine  operated  by 
compressed  air.  It  consists  of  a  rigid  frame  F  on  which  the 
air  cyUnder  C  is  mounted;  P  is  the  piston;  AB  is  the  piston 
rod  pinned  to  the  piston  at  A  so  that  the  rod  can  be  rotated 
somewhat  about  A  inside  of  the  (hollow)  piston;  the  toggle  link 
BD  is  pinned  to  the  frame  at  D;  the  toggle  Unk  BE  is  pinned 
to  the  plunger  Q  (movable  iji  a  vertical  guide  on  the  frame)  at 

E;   HE  are  the  rivet  dies  between  which  the 

rivet  is  squeezed.    AB  =  ig  ins.;  BD  =  13  ins.; 

BE  =  10  ins.;  the  diameter  of  the  cylinder  is 

10  ins.    Assume   the   air  pressure  to  be   100 

lbs/in^   and   then    determine    the  pressure  at 

the  pins  D  and  E,  the  pressure  against  the  guide, 

and  the  pressure  on  the  rivet.     (To  "lay  out " 

this  mechanism  begin  at  D,  then  fix  A,  then  B, 

and  then  E.)     Solve  the  problem  when  A  is 

advanced  2  ins.  from  the  position  shown. 


-Air 


2000\lbs. 


1 


500\lbi. 


iyuiD5.pern:       ^  V  D  4' 


Fig.  18 


Fig.  17 


3S-(i2).  The  beam  AB  (Fig.  18)  is  supported  at  C  and  D,  and  it  sustains  three 
loads  as  shown.  The  beam  weighs  50  lbs.  per  lineal  feet.  Determine  each  supporting 
force,  or  reaction. 


327 


36-(i2).  Fig.  19  represents  a  shutter  dam;  AB  is  the  shutter,  and  CD  and  CE  are 
braces.  The  shutter  and  the  braces  are  pinned 
together  at  C;  the  shutter  rests  against  an 
inclined  stop  at  A;  brace  CD  is  pinned  to  a 
bed  plate  at  D;  brace  CE  rests  against  a  bed 
socket  at  E.  The  shutter  is  4  ft.  wide  and  its 
length  AB  =  12  ft.  The  water  pressure  is 
16,000  lbs.,  and  its  "center"  is  at  F,  4  ft. 
from  A.  Determine  the  reactions  at  D  and  E 
due  to  the  water  pressure. 

37~(i3)-  Fig.  20  represents  a  truss  supported 
by  a  shelf  5  on  a  wall  and  a  horizontal  tie  A ; 
AB  =  g  ft.  and  BC  =  12  ft.  Determine  the 
reactions  at  A  and  B  due  to  the  loads. 

38-(i3).  AB  (Fig.  21)  is  a  beam  supported 
by  a  rod  CD  and  a  pin  a.t  A;  AB  =  git.  AC  =  3 

ft,  AD  =  8  ft., 
400  lbs.  and  the 


lOGO/is. 


U-2'6 


Fig.  19 

and  ^E  =  5  ft.    The  beam  weighs 

Determine 


load,  =  W  1000  lbs. 
the  pull  at  C  and  the  pressure  at  A. 

39-(i3).  The  crane  represented  in  Fig.  22  is  sup- 
ported by  two  floors  as  shown.  £  is  a  hole  in  the 
upper  floor  and  F  is  a  cylindrical  socket  in  the  lower 
floor.  The  crane  weighs  5  tons  and  its  center  of 
gravity  is  2  ft.  to  the  left  of  the  axis  of  the  post. 
Determine  the  pressures  on  the  floors  when  the  load 
W  is  5000  lbs. 
40-(i3).   A   and  B  (Fig.  23)  are  two  horizontal 

pegs  in  a  wall;  they  are  3  and  6  ft.  above  the  floor 

respectively,  and   the   horizontal   distance    between 

them  is  4  ft.     A  smooth  straight  bar  CD,  15  ft.  long 

and  weighing  200  lbs.,  is  placed  under  A  and  over  B 

with  its  lower  end  on  the  floor,  but  is  not  sprung  into 

that  position.     Determine  all  the  pressures  on  the 

bar,  due  to  its  own  weight. 

41.  AB  (Fig.  24)  is  a  bar  12  ft.  long  fastened 
to  the  floor  at  ^  by  a  pin  and  it  rests  at  C  on  a 
smooth  cylinder  4  ft.  in  diameter.  The  center 
of  the  cylinder  is  6  ft.  to  the  right  of  A  and  is 
connected  by  a  horizontal  cord  to  the  bar  at  D. 
A  weight  of  100  lbs.  is  hung  on  the  free  end  of 
the  bar.  What  is  the  pressure  between  the 
bar  and  the  cylinder;  between  the  cyUnder 
and  the  floor;  what  is  the  tension  in  the  cord; 

j.; 2o'- >i      ^^^  what  is  the  pressure  exerted  by  the  pin  on 

the  bar  A  ?     Consider  the  cyUnder  and  the  bar 
as  weightless. 

42.  AB  (Fig.  25)  is  a  bar  20  ins.  long,  and 
weighs  10  lbs.    It  rests  on  a  peg  C  and  against  a 


?vj 


.jf._. 


Fig.  22 


328 

smooth  wall  at  ^ ,  as  shown.     What  vertical  force  applied  at  B  will  preserve  the 
equilibrium  of  the  bar? 

43.    If  the  weight  of  the  bar  in  Prob.  42  is  12  lbs.  and  a  load  weighing  4  lbs.  is 
suspended  at  B,  at  what  angle  must  the  bar  be  placed  to  insure  equilibrium? 


/77777777777777777777P777777? 

Fig.  23 


Fig.  24. 


Fig.  25 


44-(i4).  Figs.  26  and  27  are  two  outline  views  of  a  steam  shovel;  the  former  repre- 
sents a  dumping  and  the  latter  a  digging  position.  A  is  the  "^-frame,"  B  the  boom, 
and  D  the  dipper.  The  pin  P  (axis  perpendicular  to  the  paper)  is  seated  on  the  upper 
half  of  a  "fifth  wheel  "  which  permits  swinging  of  the  boom  about  the  vertical  axis 
FQ.  Two  engines  are  mounted  on  the  boom,  —  the  main  engine  which  operates  the 
hoisting  drum,  and  the  thrusting  engine  which  operates  the  pinion  meshing  with  a 
rack  on  the  bottom  of  the  dipper  handle. 


•i/f|<-3'6"*l*--#'6"-*' 

Fig.  26 


Fig.  27 


Many  of  the  parts  of  a  shovel  are  most  severely  stressed  when  the  dipper  is  en- 
countering an  unyielding  obstruction  in  the  bank.  We  indicate  how  some  of  these 
stresses  may  be  determined  and  then  ask  the  student  to  determine  others. 

The  actual  resistance  of  the  bank  against  the  dipper  cannot  be  determined  with 
certainty  because  the  line  of  action  of  the  resistance  is  generally  unknown.  It  doubt- 
less depends  largely  on  the  direction  in  which  the  cutting  edge  of  the  dipper  tends  to 
move  in  the  bank,  determined  mainly  by  the  pull  of  the  hoisting  rope  and  the  thrust 
on  the  dipper  handle.  Some  designers  assume  that  the  hne  of  action  of  the  resist- 
ance for  the  digging  position  shown  in  Fig.  27  is  about  along  the  bottom  of  the  dipper. 
Making  this  assumption  and  analyzing  the  system  of  forces  acting  on  the  dipper  and 


329 


its  handle  (resistance  of  the  bank,  hoisting  pull,  weight  of  dipper  and  handle,  and 
thrust  on  the  handle)  we  find  that  the  resistance  is  about  20,000  lbs.  We  might 
proceed  now  and  determine  the  pressures  developed  at  various  points  in  the  structure 
and  mechanism  on  account  of  this  bank  resistance.  For  instance,  analysis  of  all 
the  external  forces  acting  on  the  boom,  dipper  and  handle,  main  and  thrusting  engines 
(resistance  of  bank,  pull  of  front  guys  G',  pin  pressure  at  base  of  boom,  and  weight 
of  parts  under  consideration)  shows  that  the  pull  of  the  guys  is  about  22,000  lbs. 

The  student  should  now  determine  the  stress  in  each  leg  of  the  /I -frame  and  that 
in  each  back  guy  G".  (These  guys  are  fastened  to  the  car  at  points  9  ft.  apart. 
22 1  ft.  from  the  base  of  the  ^ -frame  and  on  the  same  level  with  that  base.) 

45-(i4).  Suppose  that  the  shovel  is  digging  as  shown  in  Fig.  27,  but  with  the  boom 
at  right  angles  to  the  track  as  shown  in  Fig.  26.     The  pull  of  the  front  guys  is  22,000 

s.  as  in  the  preceding  problem.  Determine  the  stress  in  each  leg  of  the  /1-frame, 
and  the  stress  in  each  back  guy. 


\ 


Fig.  28 


Fig.  29 


46-(i5).   The  truss  represented  in  Fig.  28  is  supported  at  A  and  D;  CE  =  12  ft., 
Pi  =  1000  lbs.  and  P2  =  2000  lbs.    Deter- 
mine the  amount  and  kind  of  stress  in 
each  member. 

47~(i5)-  The  truss  represented  in  Fig. 

29  is  supported  at  F  and  D;  BF  =  CE 
=  12  ft..  Pi  =  P2  =  2000  lbs.,  and  P3  =  P4 
=  1000  lbs.  Determine  the  amount  and 
kind  of  stress  in  each  member. 

48-(i5).  The  truss  represented  in  Fig 

30  is  supported  at  A  and  E;  each  load 
P  =  1000  lbs 


Fig.  30 
Determine  the  amount  and  kind  of  stress  in  each  member. 


Fig.  31 
49-(i5).  The  truss  represented  in  Fig.  31  is  supported  at  each  end;  span  =  80  ft. 


330 


Fig.  32 


I,  2,  3,  6  and  the  points  3,  4,  5, 
vertices  of  parallelograms.      Draw  a  stress  dia- 
gram for  the  truss  loaded  as  shown,  and  make  a 
record  of  the  stresses  in  the  members. 

52-(i6).    Solve  Prob.  47  graphically. 

53-(i7).  Fig.  ;^^  represents  a  crane  consisting 
of  three  members,  a  boom  AC,  a.  brace  AD,  and 
a  post  BF.  The  crane  is  supported  at  E  and  F 
by  two  floors.  The  load  W=  5  tons.  Determine 
all  forces  acting  on  each  member. 

54-(i7).  The  crane  represented  in  Fig.  34 
consists  of  a  post  AB,  a.  boom  CD,  and  braces 
DE  and  FG.  The  crane  is  supported  by  sockets 
at  A  and  B  as  shown.  The  boom  passes  freely 
through  a  smooth  slot  in  the  post  at  H  so  that 


and  rise  =  20  ft. ;  consecutive  points  on 
AG  are  equidistant;  DI  is  perpendicular 
to  ^G;  H  bisects  AI,  and  /  bisects  GI; 
each  load  =  1000  lbs.  Determine  the 
amount  and  kind  of  stress  in  IK,  and  be 
prepared  to  describe  how  to  determine 
the  stress  in  every  other  member. 

5o-(i6).    Solve  Prob.  46  graphically. 

Si-(i6).  The  truss  represented  in  Fig. 
32  is  supported  at  each  end.  The  points 
7  are  at  the 


-  II' 


A 


— -;??9Z-^ 


B 


D 


-*r- 


^ 


w 


Floor     ^1 


I 

_]^  \j  Floor    J;_ 


Fig.  33 


any  reaction  existing  there  will  be 

xC  vertical.      The  counterweight  at 

O  is  I  ton,  the  load  W  is  \  ton, 

and  the  latter  is  21  ft.  from  the 

axis  of  the  post.     Determine  all 

the  forces  which  act  upon  each 

member. 

55-(i7).   Fig.   35    represents    a    certain    type    of 

hydrauhc    crane.     It    consists   of   a    post    AB,    an 

hydraulic  cylinder  C  mounted  on  the  post,  a  large  sleeve  5" 


m 


B 


Fig.  34 


which  can  be  slipped 
along  the  post,  two  rol- 
lers D  and  E  mounted 
on  the  sleeve,  a  boom 
EF,  and  a  tie  rod  FG.  When  water  (under 
pressure)  is  admitted  to  the  cylinder,  the 
pistons  are  pushed  upward;  the  upper  one 
bears  against  the  sleeve,  and  rolls  the  entire 
part  DEFG  up  along  the  post.  Let  the  load 
W=  2  tons  and  suppose  that  it  is  10  ft.  out 
from  the  axis  of  the  post;  then  determine  all 
the  forces  which  act  upon  each  pin  {D,  E, 
and  G). 


Fig.  35 


331 


56-(i7).  Solve  Prob.  54  but  take  into  account  the  weights  of  the  members  of  the 
crane  as  follows:  post  AB  =  0.7  ton,  boom  CD  =  0.5  ton,  brace  DE  =  0.3  ton,  and 
brace  FG  =  0.6  ton.     Middle  of  boom  is  5  ft.  6  ins.  from  axis  of  post. 

57-(i7).  Solve  Prob.  53  but  take  into  account  the  weights  of  the  members  which 
are  as  follows:  post  BF=  0.5  ton,  brace  AD  =  0.2  ton,  and  boom  AC  =  0.7  ton.  The 
boom  is  18  ft.  long;  its  center  of  gravity  is  2  ft.  6  ins.  from  B. 

58-(i8).  Fig.  36  represents  a  crane  supported  by  a  foot-step  bearing  at  B  and  a 
collar-bearing  at  C.     B  can  furnish  horizontal  and  vertical  support,  and  C  can  furnish 


-^^^''' 


6^ >-To  Hoist 

Fig.  36 


Fig.  37 


horizontal  support  only.  The  pulleys  E  and  F  are  i  ft.  in  diameter;  the  noisting 
cable  enters  the  post  at  F,  descends  through  the  post,  over  pulley  G,  and  to  the 
hoist  as  shown.  The  counter-weight  H  is  2  tons  and  the  load  4  tons.  Determine  all 
the  forces  which  act  upon  each  member. 

S9-(i8).  The  crane  represented  in  Fig.  37  consists  of  a  post  AB,  a  boom  CD,  and  a 
tie  rod  DE.  The  pulley  at  D  and  the  winding  drum  at  G  are  i  ft.  in  diameter.  The 
load  W  is  I  ton.    DE  =  12  ft.    Determine  all  the  forces  which  act  on  each  member. 

6o-(i8).  Imagine  the  winding  drum  (Prob.  59)  to  be  mounted  in  bearings  at  H 
(supported  by  the  brace  CD)  instead  of  at  G.     Then  solve. 


B 


c 


A  Cord 

t 


B 


mmmmmm 


cc 


wmmm/// 

B 


C 
Fig.  38  Fig.  39  Fig.  40  Fig.  41 

6i-(i9).  A  (Fig.  38)  weighs  100  lbs.  and  B  200  lbs.  A,  B,  and  C  are  very  rough. 
Make  separate  sketches  of  A  and  B  and  represent  all  the  forces  which  act  on  each 
body  when  P  =  20  lbs.  (not  large  enough  to  produce  any  shpping). 

62-(i9).  A  (Fig.  39)  weighs  100  lbs.  and  B  200  lbs.  For  A  and  B,  ft  =  I;  for 
B  and  C,  fi  =  f.    How  large  must  P  be  to  cause  shpping? 

63~(i9)-  -4  (Fig.  40)  weighs  100  lbs.;  the  surfaces  in  contact  are  very  rough; 
P  =  50  lbs.,  and  a  =  20°.     Determine  the  friction  F  and  the  normal  pressure  N. 

64-(i9).  A  (Fig.  40)  weighs  100  lbs.;  a  =  20°,  and  /x  =  0.6.  How  large  must 
P  be  to  start  A  ?     How  large  is  F  when  slipping  impends? 

65-(i9).  A  (Fig.  40)  weighs  100  lbs.,  a  =  40°,  /j  =  0.6,  P  =  200  lbs.  Does 
P  move  A  ? 

66-(2o).    Same  as  Prob.  63  but  refer  to  Fig.  41. 


Same  as  Prob.  64  but  refer  to  Fig.  41. 
Same  as  Prob.  65  but  refer  to  Fig.  41, 

Fig.  42  represents  a  double-wedge  device  for  raising  and  lowering  a 
heavy  load  W*  The  device  consists  of  wedges  A  and  B 
and  bearing  blocks  C  and  D;  W  =  200,000  lbs.  The 
coefficient  of  friction  is  0.5.  How  large  are  the  required 
pushes  P  to  raise  the  load?  How  large  are  the  required 
pulls  to  lower  the  load?  (First  consider  C  and  determine 

jX.-^w,M^^^,„.Z,v?w.jj^,^^^v^^y    the  forces  acting  upon  it.) 

7o-(2o).   Fig.  43  represents,  somewhat  conventionalized, 
■  ^^  an  adjusting  device  used  in  making  the  closure  (insertion 

of  the  last  few  members)  of  a  large  cantilever  bridge  (Beaver  River)  .f   The  mechanical 

elements  are  a  double  wedge  IF,  a  screw  5",  and  a  lever  L.     The  accessories  are  a  head 

piece  ZT,  two  struts  A,  and  two  wedge-blocks  B\  they  are 

pin-connected  as  shown.     C  and  C  are  two  portions  of  the 

bridge  member  to  be  connected;  they  are  under  compres- 
sion  P  and  pin-bear  against  the  wedge  blocks  B.     The 

nut,  which  bears  against  the  head  piece,  can  be  turned  by 

means  of  the  lever,  and  the  screw  and  wedge  raised  or 

lowered.     Raising  the  wedge  separates  the  wedge  blocks 

and  parts  C  and  C     Determine  the  necessary  moment  (of 

force)  on  the  lever  for  raising  the  wedge  against  pressures 

P  =  1,235,000  lbs.,  assuming  that  the  struts  A  are  vertical 

and  the  following   data:    mean  diameter  of  screw  =  4^ 

ins.;  pitch  of  screw  =  I  in.;   bevel  of  wedge  (each  side) 


Fig.  43 


=  I  in  10;  mean  radius  of  nut  where  it  bears  on  the  head 

piece=9  ins.;  coefficient  of  friction  for  all  rubbing  surfaces 

=  J.     (Consider  first  a  wedge-block,  and  determine  all  the  forces  which  act  upon  it.) 

7i-(2o).  Fig.  44  represents  a  screw 
toggle  used  in  the  erection  of  a  steel  arch 
(Niagara  Falls  and  Clifton  Bridge). t  It 
consists  of  four  multiple  links  pinned  to- 
gether as  shown,  a  right-and-left  screw  5 
with  nuts  N,  and  a  lever  L.  The  toggle 
is  supported  by  the  anchor  rod  R  and  brace 
B.  The  "pulling  end  "  of  the  toggle  was 
connected  to  the  arch  under  construction, 
supplying  the  supporting  force  P.  Assume 
mean  diameter  of  screw  =  2  ins.,  pitch 
=  5  in.,  coefficient  of  friction  =  0.3 ;  also 
that  now  the  diagonal  MM  =  16  ft.,  4  ins., 
and  the  diagonal  NN  =  4  ft.,  4  ins.     De- 


FlG.  44 


termine  the  couple  on  the  lever  which  will  shorten  NN;  which  will  lengthen  NN. 

72-(2o).    Solve  Prob.  34,  taking  into  account  the  friction  at  all  rubbing  surfaces 
(pins,  piston,  and  guide).    Pins  A  and  E  are  2  ins.  and  pins  B  and  D  are  3  ins.  in 


*  Engifieering  News,  July '15,  1911.  f  Engineering  Record,  June  10,  1911. 

t  Skinner's  Details  of  Bridge  Construction. 


333 


diameter.     The  coefficient  of  friction  is  i.     (Solve  graphically  and  make  drawing  of 
riveter  full  size  or  larger.) 

73-(2o).   Fig.  45  represents  a  band-brake.     The  diameter  of  the  wheel  is  i  ft.,  8 
ins.,  the  angle  of  lap  =  255°,  P  =  60  lbs.,  and  the  coefficient 
of  friction  is  i;  the  wheel  is  turning  clockwise.     Compute  the 
fractional  moment  and  the  pull  on  the  pins  A  and  B.     Solve  for 
the  case  when  the  wheel  is  turning  in  the  other  direction. 

74-(2i).  Fig.  46  represents  a  crank-arm  for  a  shaft,  by 
plan  and  elevation  —  dotted  Unes  to  be  disregarded.  Locate 
the  center  of  gravity  of  the  arm. 

75-(2i).  Solve  Prob.  74  but  change  width  at  thin  end  as 
shown  by  dotted  lines.  (See  Obelisk,  Art.  24.) 

76-(2i).   Fig.  47  represents  a  connecting  rod  for  a  steam 
engine  by  plan  and  elevation 
the  distances  of  the  center  of  gravity  from  the  center  of  each  hole. 


1 4'--     - 

Fig.  45 

The  rod  is  i  J  ins.  thick  except  as  noted.     Determine 


Jl. 

i_ 


[<.-4"-H< 


■7"-->k-J"-H 


a- 

7: 


el 


CM 


~CM. 


■MCM 

CM 


— 

T 

• 

"tn 

i 

I         t"'.     /  "1 


/3' 


-^,//'K2'H 


cvi 

J: 


-■X 
CM 


Fig.  46 


Fig.  47 


77-(2i).  Fig.  48  represents  a  thin  plate  into  which  holes  were  punched  at  A  and 
B,  and  the  pieces  glued  on  at  C  and  D  respectively.  Area  of  hole  A  =  4  ins.^;  that 
of  5  =  2  ins^.     Locate  the  center  of  gravity  of  the  modified  plate. 


Fig.  48 


K"  4" ->i  !<■•  4"->i 


»o 


■-inj 


1 


'_¥ ^ 


Fig.  50 


78-(2i).  Fig.  49  represents  a  five-sided  prism  made  from  a  rectangular  prism  by 
bevelling  one  edge  as  shown.  There  is  a  cylindrical  hole  2  ins.  in  diameter  in  the 
piece,  its  axis  being  parallel  to  the  line  AB,  4  ins.  from  the  face  ABODE,  and  i^  m. 
below  the  face  ABG.     Find  the  center  of  gravity  of  the  modified  prism. 

79-(2  2).  Fig.  50  is  the  cross  section  of  a  steel  beam  "built- 
up  "  of  two  angles  5X4XI  ins.  and  a  plate  8  X  f  in.  The 
centroid  of  each  angle  is  1.57  ins.  from  the  back  of  the  shorter 
leg.  Determine  the  position  of  the  centroid  of  the  entire  section. 
8o-(2  2).  Fig.  51  represents  the  cross  section  of  a  machine 
part.  Determine  the  position  of  the  centroid  of  the  crosssection. 
8i-(23).  Prove  that  the  distance  of  the  centroid  of  a  triangle 
from  its  base  equals  one-third  the  altitude. 

82-(23).  Prove  that  the  distance  from  the  centroid  to  the 
base  of  a  paraboloid  of  rr  volution  formed  by  revolving  a  pa- 
rabola about  its  axis  equals  one-third  the  altitude  a  (see  Fig. 
168,  page  94). 


334 

83-(24).  Prove  construction  (i),  page  98,  for  locating  the  centroid  of  a  trapezoid; 
also  construction  (2),  page  98,  for  locating  the  centroid  of  any  quadrilateral. 

84-(2  5).  A  cord  is  supported  at  two  points  on  the  same  level  30  ft.  apart,  and  its 
lowest  point  is  8  ft.  below  the  level  of  the  supports.  If  the  load  is  20  lbs.  per  hori- 
zontal ft.,  what  are  the  tensions  at  the  supports  and  at  the  lowest  point? 

85-(26).  A  cable  is  to  be  suspended  between  two  points  at  the  same  level  200  ft. 
apart;  the  sag  is  to  be  80  feet.     Determine  the  length  of  the  cable. 

86-(27).  A  rope  100  ft.  long  is  suspended  from  two  points  A  and  B  at  the  same 
level  80  ft.  apart.  A  body  weighing  1000  lbs.  is  suspended  from  a  point  Cx  ft.  dis- 
tant from  A.  Determine  the  tension  in  AC  when  x  =  20,  30,  40,  50,  60,  70,  and  80 
ft.     Make  a  graph  showing  how  the  tension  varies  with  x. 


87-(28).    Fig.  52  is  a  chronographic  record  of  the  launching  of  the  U.S.S.  Cali- 
fornia {Transactions  of  Naval  Architects  and  Marine  Engineers,  Vol.  12).     Determine 


FINISH 

74  75  Seconds. 


/  Revolution  of  Chronograph  CyfMer  iv  40-ft.  Travel  of  Ship 

(Time  between  Notches  =  g6econd } 

Fig.  52 

the  velocity  of  the  ship  at  the  twentieth  second  in  the  following  three  ways:    first, 
from  the  average  velocities  for  at  least  four  intervals  after  the  instant ;   second,  from 
the  average  velocities  for  at  least  four  intervals  before  the  instant;   third,  from  the 
average  velocities  for  the  half-seconds  immediately  before  and  after  the  instant. 
88-(28).   The  following  velocities  (feet  per  second)  were  computed  from  the  chron- 


335 

ographic  record  (Fig.  52)  by  taking  the  mean  of  the  average  velocities  for  the  half- 
seconds  immediately  preceding  and  following  the  instants  or  times  listed  below. 

t=  15    16    17    18    19    20    21    22    23    24 

v=      2.50  3.00  3.55  4.20  4.80  5.45  6.10  6.75  7.45  8.15 

Compute  the  acceleration  for  /  =  16  sees. 

89-(28).  Reduce  a  sprint  of  100  yds.  in  10  sees,  to  miles  per  hour.  Compare  the 
retardation  of  a  train  at  4  mi/hr/sec  with  the  retardation  of  gravity  on  a  ball  thrown 
vertically  upward. 

90-(28) .  A  point  P  moves  in  a  straight  line  so  that  s  =  2fi  —  ^f,  where  5  (in  feet) 
equals  the  distance  of  P  from  a  fixed  origin  in  the  path  at  any  time  /  (in  minutes). 
Determine  the  velocity  and  acceleration  when  /  =  i  min.;  when  t  =  2  mins.  Inter- 
pret the  negative  signs. 

9i-(28).  A  certain  point  P  of  a  mechanism  is  made  to  move  in  a  straight  Une  by 
means  of  a  crank  in  such  a  way  that  5  =  3  cos  2  (9,  where  5  =  the  (varying)  distance 
in  feet  of  P  from  a  fixed  origin  in  the  path  of  P,  and  e  =  the  (varying)  angle  which 
the  crank  makes  with  a  fixed  line  of  reference.  The  crank  rotates  uniformly  at 
100  rev/min.  Determine  position,  velocity,  and  acceleration  of  P  when  e  =  60°. 
Interpret  signs  of  the  results. 

92-(28).  In  a  certain  "gunnery  experiment"  the  shot  was  fired  through  screens 
placed  150  ft.  apart.  The  times  (in  seconds)  of  piercing  were  observed  with  the 
following  results: 


screen 

I 

2 

3 

4 

5 

6 

7 

time 

0 

0 . 0666 

o- 1343 

0.2031 

0. 2729 

0-3439 

0.4161 

Determine  the  velocity  at  the  fourth  screen. 

93-(28).  A  point  P  moves  in  a  straight  line  so  that  0  =  4—2/,  where  a  is  in 
feet  per  minute  per  minute  and  t  in  minutes.  When  t  =  o,  v  =  o  and  s  =  o.  De- 
termine general  formulas  for  v  and  s.     What  are  v  and  5  when  /  =  4  ?  when  /  =  5  ? 

94-(28).  A  certain  electric  train  can  get  up  full  speed  of  24  mi/hr  in  a  distance  of 
150  ft.,  and  can  stop  from  full  speed  in  a  distance  of  100  ft.  What  is  the  shortest 
time  in  minutes  in  which  the  train  can  make  a  run  between  two  stations  650  ft. 
apart,  the  train  starting  from  one  station  and  coming  to  full  stop  at  the  other  ? 
(Assume  that  the  starting  and  stopping  are  accomplished  uniformly  with  respect  to 
time.) 

95-(28).  A  certain  train  can  be  retarded  at  a  rate  of  4  mi/hr/sec  by  braking. 
Determine  the  times  (in  seconds)  and  the  distances  (in  feet)  in  which  the  train  can 
be  stopped  from  10,  20,  30,  and  40  mi/hr.  (Assume  that  the  retardation  is  the  same 
at  all  speeds.) 

96-(29).  Draw  the  distance-time  and  velocity-time  graphs  for  the  interval  from 
15  to  24  sees,  of  the  launching  mentioned  in  Prob.  87,  and  determine  the  velocity 
and  acceleration  at  the  twentieth  second  from  the  graphs. 

97-(29).  Fig.  53  shows  the  acceleration-time  graph  for  a  certain  rectilinear  motion. 
When  t  =  o,v  and  s  =  o.     Construct  the  v-t  and  s-l  graphs. 

98-(29).  Make  a  sketch  of  the  velocity-time  graph  for  the  train-run  described  in 
Prob.  94,  calling  the  lengths  of  the  three  periods  /i,  /2,  and  h  respectively.  Then  use 
the  principle  that  "  area  under  the  curve  "  represents  distance  travelled  to  find 
values  of  h,  t^,  and  (3,  and  finally  the  time  for  the  entire  run. 


336 


99~(3o)-  The  period  of  a  certain  simple  harmonic  motion  is  8  sees.,  and  the  ampli- 
tude is  6  ins.  What  is  the  maximum  velocity  ?  the  maximum  acceleration  ?  For 
the  motion  from  one  extreme  point  in  the  path  to  the  center,  what  is  the  average 
velocity?   the  average  acceleration? 

ioo-(3o).  Four  particles,  Qi,  Qi,  Q3,  and  Q^,  are  describing  simple  harmonic  motions 
in  AB  (Fig.  54);   the  period  of  each  motion  is  8  sees.     At  a  certain  instant  the  four 


5  sees. 


B 


Y    3 


X 


B 


>/////////>////////r'f^i'^w/MW////w, 


Fig.  53 


Fig.  54 


Fig.  55 


particles  are  at  points  i,  2,  3,  and  4  respectively;  Qy  and  ^3  are  moving  toward  the 
right  and  Q^  and  Q^^  are  moving  toward  the  left.  Write  out  the  expressions  for  the 
X  coordinates  of  the  moving  points  /  sees,  after  the  instant  mentioned.  {AB  =.12 
ins.,  and  is  divided  into  sixths  by  the  points.) 

ioi-(3i).  A  (Fig.  55)  weighs  200  lbs.,  B  weighs  100  lbs.;  the  coefficient  of  friction 
"  under  "  .-1  is  \,  that  under  5  is  };  P  =  300  lbs.  Determine  the  acceleration  of  -4 
and  B,  and  the  tension  in  the  rope  connecting  them. 


B 


•»r/j/>///^//j////'/////, 


Fig.  56 


1 


I       B      I 


''mTTTTTTTTT^TTTTTTTmr. 

Fig.  57 


B 


•'WW/Wr  '^M////U- 


X'- 
Fig.  58 


i02-(3i).  Suppose  that  the  supporting  surface  in  the  preceding  problem  is  not 
horizontal  but  inclined  at  30  degrees  to  the  horizontal.     Then  solve. 

io3-(3i).  A  (Fig.  56)  weighs  50  lbs.  and  B  weighs  100  lbs.;  the  pull  P  gives  A 
and  B  an  acceleration  of  2  ft/sec/sec.  Determine  the  magnitude  and  direction 
(referred  to  the  horizontal)  of  the  pressure  between  A  and  B. 

io4-(3i).  Two  bodies  are  connected  somewhat  as  two  cars,  and  are  placed  on  a 
plane  inclined  at  30  degrees  to  the  horizontal.  The  lower  one  weighs  600  lbs.  and  is 
smooth,  that  is,  there  is  no  resistance  to  its  sliding  on  the  plane.  The  upper  one  weighs 
1000  lbs.,  and  the  coefficient  friction  under  it  is  y"^.  With  what  acceleration  will 
the  bodies  slide  down  when  released  ?  WiU  there  be  tension  or  pressure  at  the  con- 
nection ?     What  is  its  value  ? 

io5-(3i).  The  weights  of  .1,  B,  and  C  (Fig.  57)  are  50,  100,  and  200  lbs.  respec- 
tively. Contacts  between  .4,  B,  and  C  are  very  rough;  between  C  and  D  very 
smooth;  P  =  100  lbs.  Determine  the  forces  which  the  bodies  exert  upon  each 
other.     Sketch  each  body  separately,  showing  the  forces  acting  on  it. 

io6-(3i).  A  (Fig.  58)  weighs  100  lbs.,  and  B  weighs  200  lbs.  The  coefficient  of 
kinetic  friction  under  -B  is  |;  the  coefficient  of  static  friction  under  A  is  ■^\.  When 
P  =  75  lbs.,  will  A  slip?  How  great  is  the  friction  under  .4?  How  large  a  force 
P  would- just  make  .4  slip? 

io7-(3i).  A  (Fig.  56)  weighs  50  lbs.,  and  B  weighs  100  lbs.  C  is  perfectly  smooth; 
the  coefl&cient  of  static  friction  "between"  A  and  .B  is  ^;   the  angle  between  the 


337 

top  of  B  and  the  horizontal  is  25  degrees.  How  great  may  P  be  without  making  A 
slip  on  5  ? 

io8-(3i).  A  (Fig.  59)  weighs  100  lbs.,  and  B  weighs  50  lbs.  The  coefficient  of 
friction  under  A  is  \.  Neglect  the  inertia  of  the  pulley  and  the  friction  at  its 
axle,  and  find  the  acceleration  of  A  and  B,  and  the  tension  in  the  cord.  (The  re- 
sultant of  the  three  forces  acting  on  A  is  T—  20,  where  T  =  tension;  and  the  resultant 
of  the  two  forces  acting  on  5  is  50  —  T.  Now  write  the  equations  of  motion, 
R  =  {W/g)a,  for  A  and  B,  and  solve  them  simultaneously  for  a  and  T.) 

io9-(3i).  Show  that  the  acceleration  of  the  suspended  bodies  and  the  tension  in 
the  cord  of  the  Atwood  machine  (Fig.  60)  are  respectively 

(2  W1W2) 


(iK-Il)      and      T 


Wi  +  IF2 


Wi^Wi 


when  the  inertia  of  the  pulley  and  the  axle  friction  are  negligible. 

iio-(3i).    Fig.  61  represents  a  simple  engine,  without  connecting  rod.     Stroke  = 
18  ins:,  speed  =  150  r.p.m.     Piston  and  rod  weigh  120  lbs.     When  x  =  3  ins.,  steam 


A 


'^w//mhr^[ 


i    ' 
B 


Fig.  59 


Fig.  60 


Fig.  61 


M 


pressure  =  2000  lbs.     Determine  the  pressure  of  crank  pin  P  on  the  piston  rod. 
When  the  piston  is  advanced  6  ins.  beyond  the  position  shown  (x  =  —  3  ins.),  the 
steam  pressure  is  still  2000  lbs.     Determine  the  pressure  of  the  pin  on 
the  piston  rod  for  this  position. 

iii-(3i).  Suppose  that  Fig.  61  represents  an  air  compressor,  steam 
being  changed  to  air  and  the  crank  turning  clockwise.  Determine 
the  pressures  of  the  crank  pin  for  the  two  positions  mentioned  in  the 
preceding  problem  (x  =+  3  ins.  and  x  =—  3  ins.). 

ii2-(3i).  Fig.  62  represents,  in  prmciple,  a  certain  "throw  "  test- 
ing machine  for  subjecting  a  metal  specimen  to  rapid  changes  of 
direct  stress  (tension  and  compression).  S  is  the  specimen,  firmly 
screwed  into  two  bosses  M  and  N.  W  is  a  weight  firmly  fastened 
to  the  lower  boss.  The  parts  named  can  be  oscillated  in  the  vertical 
guides  G  by  means  of  an  ordinary  crank-connecting  rod  mechanism 
(OP-PC).  When  the  machine  is  not  running,  the  specimen  is  sub- 
jected to  a  tension  equal  to  the  weight  of  N  and  W.  When  the 
machine  is  running,  the  stress  on  the  specimen  changes  continuously.* 
Let  OP  =  I  in.,  PC  =  g  ins.,  weight  of  N  and  W  =  25  lbs.,  and 
speed  of  crank  =  2000  rev/min.  Determine  the  stress  on  the  spec- 
imen at  each  end  of  a  stroke  or  oscillation,   and  at  the  middle  of  the  stroke. 

ii3-(3i)-    Take  data  except  speed  as  in  preceding  problem.     Determine  the  speed 

*  For  detailed  description  see  Phil.  Trans.  Roy.  Soc,  Ser.  A,  Vol.  199  (1902). 


g-In^g 


1 


Fig.  62 


338 


which  would  make  the  stress  on  the  specimen  equal  to  zero  at  the  upper  end  of  the 
stroke.     What  would  the  stress  be  at  the  lower  end  at  that  speed  ? 

ii4-(32).  A  point  P  starts  at  A  (Fig.  63),  and  moves  in  the  circle  as  indicated 
traversing  distance  s  so  that  s  =  2  fi,  where  t  is  time  after  starting  in  seconds  and  5 
is  in  feet;  radius  OA  =  20  ft.  Draw  the  hodograph  for  the  first  3  sees.  Then  de- 
termine the  average  accelerations  for  the  intervals  i  to  3,  1.5  to  3,  2  to  3,  2.5  to  3. 
Next  determine  the  magnitude  and  direction  of  the  acceleration  when  t  =  3  from 
these  average  accelerations. 

Y 


Y 

f 

/  \ 

A 

M 

y 

)-^ 

Fig.  63 


Fig.  64 


ii5-(33).  Determine  Vx,  Vy,  v,  Ox,  Oy,  at,  an,  and  a  for  the  motion  described  in 
Prob.  114  and  when  /  =  3  sees. 

ii6-(33).  The  point  Q  (Fig.  64)  on  the  rim  of  a  wheel  rolling  in  a  straight  line 
describes  a  curve  known  as  cycloid.  Let  v'  =  velocity  of  the  center  of  the  wheel  C, 
a'  =  the  acceleration  of  C,  and  R  =  radius  of  the  wheel.  Find  formulas  for  the  x 
and  y  components  of  the  velocity  and  acceleration  of  Q  when  in  the  position  shown. 
(Let  5  =  the  abscissa  of  C,  and  x  and  y  =  the  coordinates  of  Q.  Then  x  =  s  — 
R  sin  d,  and  y  =  R  {i  —  cos  6) ;  also  5  =  Rd.) 


f  - 


Fig.  66 


ii7-(33).  A  point  Q  describes  a  simple  harmonic  motion;  the  frequency  =  100 
(to  and  fro)  oscillations  per  minute  and  the  amplitude  =  3  ft.  Determine  the 
average  accelerations  of  Q  for  the  following  distances  traversed :  first  6  ins.  from  one 
end  of  its  path;   second  6  ins.;   third  6  ins.;  and  first  18  ins. 

ii8-(33).  A  16-inch  gun  can  give  a  projectile  weighing  2400  lbs.,  a  muzzle  velocity 
of  1465  mi/hr,  and  at  an  elevation  of  15  degrees  can  throw  it  approximately  9  mi. 
Compute  the  range  for  the  velocity  and  elevation  stated  neglecting  air  resistance,  and 
compare  with  the  actual  range. 

ii9-(34).  A  cylinder  C  (Fig.  65)  is  suspended  by  a  cord  and  rests  against  a  smooth 
inclined  plane  P  as  shown.  The  cylinder  weighs  20  lbs.;  its  diameter  is  one  foot. 
The  plane  is  rotated  at  30  rev/min  about  the  vertical  axis  AB.  Determine  the 
tension  in  the  cord  and  the  pressure  against  the  plane. 

i2o-(34).   CD  (Fig.  66)  is  a  vertical  axis  about  which  E  can  be  rotated.    A  is  a 


339 


5' 


S" 


B 


V 


i^- 


D 


T      . 
■3' * 


Fig.  67 


— T 


body  resting  on  E,  and  B  is  suspended  by  means  of  a  cord  fastened  to  A  as  shown. 
A  weighs  10  lbs.  and  B  weighs  20  lbs.  Suppose  that  E  makes  30  rev/min;  then 
compute  the  pressure  at  the  stop  S.  The  centers  of  A  and  B  are  5  and  3  ft.  from 
CD  respectively.     (Neglect  friction  under  A,  a,t  B  and  the  pulley  axle.) 

i2i-(34).  Suppose  that  ,1  and  B  in  Prob.  120  are  rough,  the  coefi&cients  of  static 
friction  being  j  for  each.     What  rate  of  rotation  would  lift  B? 

i22-(34).  T  (Fig.  67)'  is  a  horizontal  whirling 
table.  A  and  B  are  spheres  connected  by  an  elastic 
cord,  the  tension  in  which  is  30  lbs.  when  the  table 
is  at  rest.  A  weighs  ir  lbs.  and  B  weighs  40  lbs. 
What  are  the  pressures  of  the  stops  S'  and  S"  against 
the  spheres  when  the  table  is  rotated  about  CD  at  20 
rev/min  ? 

i23-(35).  Suppose  that  the  floor  of  the  car  and  A  (Fig.  283,  Art.  35)  are  very 
rough  so  that  A  will  not  slip  on  the  car;  then  ascertain  how  great  an  acceleration  of 
the  car  would  result  in  tipping  of  A . 

i24-(35).  Suppose  that  the  coefBcient  of  friction  in  Prob.  123  is  i.  If  the  ap- 
plied push  on  the  car  is  gradually  increased,  thus  increasing  the  acceleration  grad- 
ually, will  .1  sUp  or  tip  eventually  ? 


Fig.  68 


Fig.  69 


2" 


W///////Ay/^^//////////////A 


Fig.  70 


-^zft^MI^ 


i2S-(35).  The  Scotch  cross-head  (Fig.  61)  described  in  Prob.  no  presses  against 
the  stuffing  box  and  on  the  cylinder  by  reason  of  the  weight  of  the  cross-head  and 
the  pressure  of  the  crank-pin  on  it.  Suppose  that  the  center  of  gravity  of  the  cross- 
head  is  1 5  ins.  from  the  center  of  the  slot,  the  center  of  the  piston  is  24  ins.  from 
the  same  point,  and  the  center  of  the  stuffing  box  is  13  ins.  from  0.  Determine  the 
pressures  mentioned  when  the  circumstances  are  as  in  Prob.  no  (steam  pressure  = 
2000  lbs.,  etc.). 

i26-(36).  Show  that  the  moment  of  inertia  of  the  slender  wire  AB  (Fig.  68) 
about  the  x-axis  is  §  Mr\\  —  (sin  a.  cos  a)/a\,  where  M  =  mass  of  the  wire. 

i27-(36).  Show  that  the  moment  of  inertia  of  a  right  circular  cone  about  its  axis  is 
I'ij  Mr'',  where  M  =  the  mass  of  the  cone  and  r  =  the  radius  of  its  base. 

i28-(36).  Show  that  the  moment  of  inertia  of  the  ring  or  torus  (Fig.  69)  about 
the  2-axis  is  M  (R^  +  5  r^),  where  M  =  the  mass  of  the  ring. 

i29-(36).  The  length  of  a  homogeneous  right  elliptic  prism  is  /,  and  the  semi- 
axes  of  its  cross  section  are  a  and  b.  Prove  that  the  radius  of  gyration  of  the 
prism  with  respect  to  a  line  through  its  center  of  gravity  parallel  to  the  axis  b  is 

i30-(36).  Fig.  70  is  a  section  of  a  cast-iron  flywheel;  there  are  six  spokes.  The 
cross  section  of  each  spoke  is  elliptical,  the  axes  of  the  ellipse  being  2  inches  and  5^ 
ins.  long.  Compute  the  moments  of  inertia  of  rim,  spokes,  and  hub  with  respect  to 
the  axis  of  the  wheel;  also  the  radius  of  gyration  of  the  wheel  about  that  axis. 


340 

i3i-(37)-  In  order  to  produce  a  tension  of  loo  lbs.  in  the  cord  of  Ex.  2,  Art.  37, 
how  heavy  must  the  suspended  body  be  ? 

i32-(37)-  A,  B,  and  C  (Fig.  71)  weigh  100  lbs.,  30  lbs.,  and  34.4  lbs.,  respectively. 
The  diameter  of  C  =  2  ft.  3  ins.,  and  the  radius  of  gyration  of  C  about  the  axis 
of  rotation  =  i  f t. ;  0  =  30  deg.  Friction  under  .1 ,  when  the  system  is  moving, 
=  ID  lbs.  Determine  the  acceleration  of  A,  B,  and  C,  and  the  tensions,  the  system 
having  started  without  initial  velocity.     (Neglect  axle  friction.) 

i33~(37)-  A,  B,  and  C  (Fig.  72)  weigh  50  lbs.,  100  lbs.,  and  150  lbs.  respectively. 
C  is  a  solid  disk  of  cast  iron  16  ins.  in  diameter.  Determine  the  acceleration  of  A, 
B,  C,  and  also  the  pulls  of  the  cord  on  A  and  B.     (Neglect  axle  friction.) 


Fig.  71 


Fig.  72 


Fig.  74 


i34-(37).  AB  (Fig.  73)  is  a  brake  for  regulating  the  descent  of  the  suspended 
body  C.  C  weighs  1000  lbs.,  the  drum  2000  lbs.,  the  diameter  of  the  drum  =12  ft., 
that  of  the  brake  wheel  =  14  ft.,  a  =  4  ft.,  6  =  6  ins.,  and  the  radius  of  gyration  of 
the  entire  rotating  system  about  the  axis  of  rotation  =  4  ft.  When  P  =  100  lbs. 
and  the  coefficient  of  brake  friction  is  |,  what  is  the  acceleration  of  C?  (Neglect 
axle  friction.) 

135^(37) •  The  wheel  A  (Fig.  74)  is  a  solid  cylinder  weighing  1000  lbs.  and  its 
diameter  is  8  ft.  It  is  desired  to  arrange  a  brake  BC  as  shown,  by  means  of  which 
the  speed  of  the  wheel  may  be  reduced  from  100  rev/min  to  zero  in  10  sees.  The 
coefficient  of  friction  at  D  =  \;  the  available  pull  P  is  100  lbs.  Determine  the 
ratio  a/h.     (Neglect  axle  friction.) 

i36-(38).  Determine  the  magnitude  and  direction  of  the  axle  reactions  in  Probs. 
131  and  132;  in  Probs.  133  and  134. 

137^(38) •   Iri  Fig.  74,  a  =  6  ft.  and  &  =  6  ins.;  the 

wheel  weighs  400  lbs.    The  coefficient  of  brake  friction 

=  |.     When  the  wheel  is  turning  clockwise,  a  push 

Pof  120  lbs.  is  appUed.     Determine  the  axle  reaction. 

i38-(38).   A  (Fig.  75)  is  a  rigid  piece  which  can  be 

rotated  about  the  vertical  axis  BC.     D  is  a  vertical 

bar  pinned   to  A    at  E,  and  rests  against  A   at  F; 

the  bar  is  14  ins.  long  and  weighs  20  lbs.     The  speed  of 

'^  rotation  is  100  rev/min.    Determine  the  pressures  on  D. 

i3g-(38).   In  Ex.  i  of  Art.  38,  §  2,  take  P  as  applied  at  F  and  solve;  then  as  appUed 

at  G  and  solve. 

i4o-(39) .  Compute  the  length  to  the  nearest  hundredth  inch  of  the  simple  seconds 
pendulum  for  your  locality. 

i4i-(4o).  The  body  C  (Fig.  76)  weighs  50  lbs.  It  is  dragged  up  the  plane  by  P 
(=40  lbs.)  and  Q  {=  20  lbs.).     The  frictional  resistance  is  5  lbs.;  a  =  30°.     Com- 


»   ■    ■      >    ,  341 

pute  the  work  done  on  C  by  each  force  acting  on  it  while  C  is  moved  from  A  to  B, 
a  distance  of  i  s  ft. 

i42-(4o).  ABC  (Fig.  77)  is  a  smooth  rail  in  the  form  of  a  vertical  semicircle  of 
4  ft.  radius.  D  is  a.  body,  weighing  50  lbs.,  which  can  be  made  to  slide  along  the 
rail.  P  is  a  force  of  150  lbs.  always  inclined  30  deg.  to  the  horizontal;  Q  is  a 
force  of  40  lbs.  always  directed  along  the  tangent.  Compute  the  work  done  on  D  by 
all  the  forces  acting  on  it  while  D  is  moved  from  .1  to  B. 


Fig.  76 


Fig.  77 


i43-(4o).  Solve  the  preceding  problem  on  the  supposition  that  P  is  always  di- 
rected toward  B. 

i44-(4o).  In  order  to  retard  the  motion  of  a  launching  ship,  ropes  were  fastened 
to  it  and  to  points  on  the  shore,  so  that  the  ship  broke  many  of  the  ropes  as  it  pro- 
gressed. In  order  to  estimate  the  retarding  effect  of  each  rope  broken,  tension  tests 
were  made  on  samples  of  the  rope  (7-in.  manilla).  Fig.  78  shows  the  average  ten- 
sion-stretch curve  for  these  tests.     The  average  strength  of  the  samples  was  about 


V 

a 

in 

0) 


«0 


^ 



^^ 

2.0 

^^ 

^ 

^ 

^^ 

y^ 

• 

y 

y 

1.0 

/ 

/ 

/ 

/ 

/ 

10000 


Pu\l, 


20000 
Pounds. 


30000 


Fig.  78 


32,500  lbs.  It  was  assumed  that  the  efficiency  of  the  knots  used  would  be  about 
80  per  cent,  and  therefore  that  the  ropes  would  fail  at  about  26,000  lbs.  On  the 
basis  of  this  assumption  and  the  curve,  it  was  estimated  that  each  rope  (20  ft.  long) 
would  do  60,000  ft-lbs.  of  work  on  the  ship  before  breaking.  Can  you  check  this 
estimate?     (Data  taken  from  Trans.  Soc.  Nav.  Archls.  and  Mar.  Engrs.,  1903,  p. 

295-) 

i45-(4i).   Show  that  the  rotational  part  of  the  kinetic  energy  of  a  rolling  sphere 

is  two-sevenths  of  its  total  kinetic  energy. 


342 

i46-(4i).   A  certain  freight  car  with  its  load  weighs  60  tons.     Each  pair  of  wheel* 
with  its  axle  weighs  1800  lbs.,  and  the  radius  of  gyration  of  a  pair  and  axle  with 
respect  to  the  axis  of  the  axle  is  0.81  ft.;  the  diameter  of  the  wheels  is  ^:^  ins.     De- 
termine the  ratio  of  the  rotational  part  of  the  kinetic 
energy  of  the  moving  car   (and  load)   to  the  trans- 
lational  part. 

i47-(42).  In  the  American  Machinist  for  Dec.  2, 
1909,  there  appears  a  communication  in  which  an 
alleged  "  favdt  in  brake  dynamometers  "  is  pointed  out 
and  explained.  The  writer  states  that  on  several 
occasions  he  got  ridiculous  results  with  a  Prony  brake. 
The  "enigma"  became  clear  to  him  when  he  en- 
countered a  "  paradox "  in  his  experimental  work, 
described  by  him  as  follows: 

In  Fig.  79  S  represents  a  shaft,  mounted  in  two  bearings 
BB',  carrying  two  levers,  arms  AA',  each  exactly  50  inches 
long  from  center  of  shaft  to  the  fulcrums  MM',  respectively, 
and  firmly  keyed  to  the  shaft.  At  K  is  represented  a  coun- 
tecweight  which  balanced  the  two  lever  arms  and  brought 
the  center  of  gravity  about  the  center  line  of  the  shaft  S; 
T  represents  a  platform  scale  and  W  represents  a  weight, 
which  weighed  100  pounds  when  placed  on  the  scales  T. 
When  W  of  loo-pounds  weight  was  hung  on  the  fulcrum 
N',  the  scales  just  balanced  at  no  pounds.  At  first  the 
paradox  almost  paralyzed  the  brain,  but  on  closer  examination  the  mystery  was  easily 
solved,  as  follows:  Considering  A  and  A'  to  be  firmly  keyed  to  the  shaft  S,  then  the  two 
arms  and  shaft  S  become  practically  one  solid  mass.  Therefore,  when  an}^  weight  IF  is  placed 
on  the  fulcrum  M' of  the  lever  A',  the  whole  mass  will  tend  to  rotate  about  a  line  passing 
through  the  points  of  support  M  and  B,  with  a  moment  of  W  times  the  lever  arm  X'.  The 
shaft  5  at  the  point  B'  will  be  fetched  forcibly  up  against  the  top  of  the  box  or  bearing  cap 
of  bearing  B',  which  will  resist  the  rotation  of  the  mass  about  MB,  with  a  balancing 
moment  equal  to  WX',  or  a  reaction  on  the  bearing  cap  equal  to 

WX'        WX 
~T^  .  or  -y-  • 


yTy^W'IOOlbs. 
Fig.  79 


Now,  it  is  evident  that  the  resultant  of  these  two  forces  is  a  downward  vertical  force  C  at 
the  point  C  equal  to  IF  +  {WX)/Y  which  load  is  distributed  between  the  points  of  support 
B  and  M  inversely  proportional  to  their  respective  distances  from  the  point  C  Hence  the 
load  on  the  scales  T  will  be  represented  by 

ZC  Z       (  WX\ 

F-I-Z'^V  +  Z  V  Y    j 

Hence  a  weight  of  100  pounds  on  the  fulcrum  M  will  produce  a  load  on  the  scales  T  equal  to 


V  +  Z 


I  ,    iooZ\ 


instead  of  100  pounds  as  generally  believed.  The  above  condition  obtains,  more  or  less,  in 
the  vast  majorities  of  dynamometers,  and  is  sometimes  so  exaggerated  as  to  make  the  re- 
sults positively  ridiculous.  In  the  case  of  a  motor  test  let  IF  represent  the  tangential  pull  on 
the  armature,  an  equal  upward  pull  on  the  opposite  side  of  the  shaft  might  tend  to  balance 
the  error,  or  it  might  tend  to  make  matters  worse,  depending  upon  the  position  of  the  other 


343 


points  in  the  diagram,  but  wherever  W  may  fall  the  results  will  be  most  erroneous.  For 
instance,  suppose  that  IF  happens  to  fall  on  the  line  MB  at  W,  then  it  is  evident  that  the 
weight  exerted  on  the  scales  T  will  be  equal  to 

^^^^,msteadof  — 
as  generally  accepted. 

Show  that  the  writer  is  mistaken  in  his  assertion  that 


(  loo  H :^ —  )  does  not  =  loo,  and  that 


WZ 

V  +  Z 


does  not  = 


NW 


v}//////7iJ/J/////////////J///////////////m777777^, 


V  +  Z 

and,  hence,  that  his  explanation  of  the  "  enigma  "  does  not  explain. 

i4o-(42).  Fig.  80  represents  Durand's  dynamometer.  A,  B,C,  and  D  are  sprocket 
wheels  of  equal  diameter;  A  and  B  are  mounted  on  a  beam  XYT  which  is  carried 
by  the  well-known  Emery  steel-plate  support  or 
knife-edge  at  E.  The  knife-edge  rests  on  the 
standard  R.  Sprocket  wheels  C  and  D  are 
mounted  on  R.  The  bars  ^^  are  fastened  rig- 
idly to  the  beam,  and  engage  loosely  with  a  pin 
on  R,  thus  limiting  rotation  of  the  beam.  The 
sprocket  chain  passes  over  A,  under  D,  over  B, 
under  C,  and  up  to  A.  The  shafts  for  C  and 
D  are  extended  forward  and  back;  and  on  these 
extensions  puUeys  may  be  mounted,  or  universal 
joint  couplings  may  be  attached,  for  the  receipt 
and  delivery  of  power.     (For  detailed   descrip-  Fig.  80 

tion  see  American  Machinist  for  June  20,  1907.) 

OEO'T  is  horizontal;  PQ  and  IH  are  vertical;  MN  and  KL  are  inclined  at  an  angle 
of  27  deg.  with  the  vertical;  OR  =  O'E  =  12  ins.;  and  ET  =  24  ins.  Suppose  that 
an  electric  motor  on  the  shaft  of  C  turns  counter-clockwise  at  100  rev/min,  and 
transmits  to  a  machine  on  the  shaft  of  D,  and  that  a  weight  of  40  lbs.  at  T  keeps 
the  beam  XY  balanced.     What  is  the  power  of  the  motor? 

i49-(42).  Assume  that  the  law  of  mean  effective  pressure  and  piston  speed  is 
represented  by  the  dotted  line  in  Fig.  331  of  the  text,  so  that 

p  =  ^o[o.9S  -  (7-^-^11,000)], 

where  p  =  mean  effective  pressure,  po  =  boiler  pressure,  and  s  =  piston  speed  in 
feet  per  minute.  Then  derive  a  formula  for  indicated  locomotive  power.  Find 
piston  speed  at  which  power  is  maximum.  Also  graph  your  formula  in  the  figure, 
calling  the  maximum  power  100  per  cent. 

i5<>-(42).  Let  D  =  diameter  of  the  driving  wheels  of  a  locomotive  in  inches; 
/  =  stroke  in  inches ;  d  =  diameter  of  the  cylinder  in  inches ;  po  =  boiler  pressure  in 
pounds  per  square  inch;  and  V  =  velocity  of  the  locomotive  in  miles  per  hour. 
Assume  that  the  mean  effective  pressure  varies  as  described  in  the  preceding 
problem.  Derive  a  formula  for  the  indicated  power  of  the  locomotive  in  horse 
powers  for  any  velocity  V. 

iSi-(43)-  A  certain  body  weighs  400  lbs.,  and  is  dragged  along  a  rough  hori- 
zontal plane  by  a  force  of  80  lbs.  The  force  is  inclined  20  deg.  upward  from  the 
horizontal;    the  coefficient  of  friction  between   the  body  and  plane  is  about  j\. 


344 


At  a  certain  point  in  the  motion,  the  velocity  of  ^  is  5  ft/sec.     What  is  the  velocity 
of  A  10  ft.  beyond  the  point? 

i52-(43).  For  the  purposes  of  comparing  the  "  running  quahties  "  of  certain  freight 
car  trucks,  they  were  tested  substantially  as  follows:  Each  one  was  made  to  roU 
down  a  steep  incline  to  give  it  "  initial  velocity,"  and  then  it  passed  onto  a  moderate 
upgrade;  the  velocity  was  measured  at  two  points  on  the  upgrade;  then  the  loss 
of  kinetic  energy  was  computed.  These  losses  furnished  a  comparison.  The  up- 
grade was  0.38  per  cent,  and  the  points  at  which  velocities  were  measured  were 
257.2  ft.  apart.  One  of  these  trucks  (four-wheeled)  weighed  18,150  lbs.;  each  pair 
of  wheels  and  axle  1800  lbs.  The  diameter  of  wheels  was  ^;i  ins.;  the  radius  of 
gyration  of  a  pair  and  axle  was  0.81  ft.  In  one  test  the  velocities  at  the  two  points 
were  14.95  ^.nd  11.05  ft/sec.  Determine  the  average  ''truck  resistance,"  a  single 
imaginary  force  equivalent  to  actual  resistances,  not  including  gravity,  on  the 
truck.      (Experiments  by  Prof.  L.  E.  Endsley  for  American  Steel  Foundries.) 

i53-(43).  The  suspended  body  C  (Fig.  81)  weighs  10  lbs.  The  coefficient  of 
friction  under  the  brake  is  ^;   n  =  4^  ins.,  ^2  =  6  ins.,  a  =  2  ft.,  and  b  =  1  it.     C 

is  allowed  to  descend  6  ft.,  thus  turning 
the  wheel,  and  then  the  brake  is  put  on, 
with  P  =  20  lbs.  How  much  farther  will 
C  descend?     (Neglect  axle  friction.) 

i54-(43).  A,  B,  and  C  respectively 
(Fig.  82)  weigh  100,  30,  and  64.5  lbs. 
The  diameter  of  C  =  30  ins.,  and  its 
radius  of  gyration  about  the  axis  of  rota- 
tion =  I  ft.;  0  =  30  deg.  The  friction 
under  A  =  10  lbs.  Determine  the  velocity  of  the  system  when  A  has  moved 
through  10  ft.  from  rest. 

i55-(43).  Copy  Fig.  339  (pertaining  to  Exs.  2  and  3,  Art.  43,  §  2)  using  scale 
I  in.  =  10,000  lbs.  and  5  mi/hr.  (a)  Make  a  graph  in  your  copy  which  will  show 
how  the  accelerating  force  is  apportioned  between  the  locomotive  and  the  cars. 
What  is  there  in  your  finished  figure  which  represents  draw-bar  pull?  (b)  Modify 
your  figure  for  the  case  of  the  train  when  on  an  upgrade  of  0.5  per  cent. 

i56-(43).  Make  graphs  showing  how  the  total  train  resistance  in  pounds  varies 
with  the  velocity  in  miles  per  hour  according  to  Schmidt's  formula  and  the  Engi- 
neering News  formula  for  the  train  described  in  Ex.  2,  Art.  43,  §  2. 

i57-(43).  Make  a  new  figure  (as  for  problem  155)  assuming  that  the  train  re- 
sistance varies  according  to  Schmidt's  formula.  First  assume  level  track;  then 
modify  the  diagram  for  the  case  of  an  upgrade  of  0.5  per  cent. 

i58-(43).  Referring  to  the  preceding  problem  with  train  on  upgrade:  (a)  Make 
a  graph  showing  how  the  acceleration  changes  with  velocity,  (h)  Find  the  time  re- 
quired for  the  velocity  to  change  from  10  to  20  mi/hr.     (See  §  3,  Art.  28.) 

i59-(43).  Make  a  graph  showing  how  the  velocity  of  the  train  of  the  preceding 
problem  (on  the  upgrade)  changes  with  the  time  (in  seconds)  during  the  run  men- 
tioned. 

i6o-(43).  Make  a  graph  showing  how  the  distance  covered  by  the  train  of  the 
preceding  problem  (on  an  upgrade)  changes  with  the  time. 


345 

i6i-(44)-  Fig-  83  represents  in  outline  a  certain  small  vertical-lift  bridge.  The 
lifting  span  is  counterweighted  as  shown.  At  the  center  of  the  span  there  is  a  cross- 
shaft  having  on  each  end  a  drum  long  enough  to  provide  for  two  up-haul  and  two 
down-haul  cables.     From  each  drum  the  cables  are  led  to  deflecting  sheaves  at  each 


Sheave 


Floor 

Spur  C5eCT/-><'ag™<  ■  Pinion 
Bevel  Gears-^^^f,  ^Drum 

Drum-        Cross  %5hafr 

Section  at  Drums. 
Sheave  Sheave 


^;;;^?p^ 


] 


heave 


<Tower 


Fig.  83 


end  of  the  span,  beyond  which  they  are  led  to  attachments  at  the  top  and  bottom  of 
the  tower,  as  shown.  The  cross-shaft  is  driven  through  a  pair  of  bevel  gears  by  a 
vertical  shaft  connected  by  a  single  set  of  spur  gearing  to  a  second  vertical  shaft  to 
the  capstan  head  of  which  the  operating  lever  is  fitted  when  the  span  is  to  be  raised  or 
lowered.     For  full  description  see  Engineering  News  for  July  18,  191 2. 

The  dimensions,  etc.,  are  as  follows,  but  those  marked  "assumed"  are  missing  in 
the  published  description:  —  Weight  of  lifting  span  =  58,000  lbs.,  and  of  each  coun- 
terweight 29,000  lbs.;  length  of  operating  lever  =  6  ft.  (assumed);  number  of  teeth 
in  pinion  17,  in  spur  gear  68,  and  in  each  bevel  gear  30;  diameter  of  drums  =  18 
ins.  (assumed);  diameter  of  (four)  deflecting  sheaves  =12  ins.,  and  diameter  of 
shafts  for  same  =  i  in.  (assmned);  diameter  of  (four)  counterweight  sheaves  = 
54  ins.  (assumed);  and  diameter  of  shafts  for  same  =  3^  ins.  (assumed);  diameter 
of  (four)  counterweight  cables  =1  in.;  and  of  other  cables  f  in.  Determine  the 
necessary  effort  (force)  at  the  end  of  the  operating  levei:  required  to  Hft  the  span; 
to  lower  it. 

i62-(44).  Fig.  84  represents  the  arrangement  of  tackle,  engines,  etc.,  used  for 
moving  a  large  building  (three  stories,  120  X  142  ft.,  weighing  about  8000  tons). 
Pulls  were  apphed  at  six  points  on  the  rear  of  the  building  as  shown.  The  four  blocks 
under  and  the  three  immediately  in  front  of  the  building  are  single  (one  sheave  or 
pulley  in  each) ;  A  and  B  are  single,  C  and  D  double,  and  E  and  F  triple.  The  pull- 
ing cable  from  each  engine  extends  to  A ,  and  is"  reeved  through  A  and  B,  ending 
at  ^;  a  second  cable  is  fastened  to  A  and  reeved  through  C  and  D,  ending  at  C;  a 
third  cable  is  fastened  to  C  and  reeved  through  E  and  F,  ending  at  E.  Blocks  E  are 
merely  hooked  to  the  three  blocks  immediately  in  front  of  the  building;  blocks  B, 
D  and  F  are  held  in  place  by  cables  fastened  to  deadmen  (buried  logs  or  the  like). 
The  runs  of  cable  from  A  to  C  and  from  C  to  £  are  really  parallel  to  the  main  runs; 


346 

they  are  shown  inclined  to  avoid  confusion  of  lines.  The  pull  of  each  engine  was 
about  one  ton.  (For  fuller  description  see  Engineering  Record  for  Nov.  22,  1913.) 
Assume  i^  to  be  1.15,  and  compute  the  total  pull  exerted  on  the  building;  also  the 
pull  exerted  on  each  deadman  cable.  Which  one  or  ones  of  all  the  cables  is  subjected 
to  the  greatest  pull? 


„ ^ 


Hohtincf^      o O 


Engine D  _ 

^■"Deadman" '      F 


Fig.  84 


i63-(44).  The  building  (preceding  problem)  was  moved  forward  40  ft.  for  each 
setting  of  the  equipment.  How  far  did  blocks  .4,  C,  and  E  travel  for  each  setting? 
How  much  cable  was  wound  on  the  drum  of  each  engine? 

i64-(45).  Fig.  85  represents  the  mechanism  for  operating  a  small  bascule  bridge 
of  a  single  draw  span.     The  train  of  gears,  A ,  B,  C,  D,  and  E  rests  on  the  (fixed)  ap- 


Quadrani" 

Keyed  or7''x 
Trunnion 


Girder 

Draw  Span 


EmO  .  NC-MQ 


16" Crank 
■■''  Handle 

Top  of  Counfer- 
weighi  and_ 
Roadway    \ 


Concrete  Girder  of 
Approach   Span 


,_<ounferweihhf  on  Meet 
f  Girders 


Fig.  85 


proach  span.  The  quadrant  and  the  draw  span  are  keyed  to  the  same  trunnion, 
supported  on  the  pier  shown.  When  the  hand  crank  is  turned  counter  clockwise 
(in  the  view  shown),  the  quadrant  rotates  clockwise,  and  the  free  end  of  the  draw 
span  lifts.  The  total  weight  of  the  draw  span  and  counterweight  is  115,000  lbs., 
and  the  center  of  gravity  of  that  (moving)  part  of  the  bridge  is  in  the  axis  of  the  trun- 
nion. The  trunnion  is  7  ins.  in  diameter.  The  following  description  of  the  gear  train 
is  sufficient  for  our  purpose: 

Gear A  B  C  D  E  Quadrant 

Number  of  teeth 13  94  15  122  11  57 


347 

(For  fuller  description  see  Engineering  News  for  July  24,  1913,  or  the  paper  by 
the  designer,  Prof.  L.  E.  Moore,  in  Engineering  and  Contracting  for  Aug.  13,  1913-) 
Determine  how  large  a  force  applied  to  the  crank  handle  at  right  angles  to  the  crank 
is  required  to  raise  the  draw  span. 

i65-(45).  Fig.  86  represents  in  plan  certain  elements  of  the  downstream  (miter) 
gate  of  the  lock  at  the  Keokuk  Dam.  Each  leaf  of  the  gate  is  hung  on  hinges  some- 
what like  an  ordinary  door;   but  the  lower  hinge  is  a  hemispherical  pivot  or  pintle 


Fig.  86 


and  ii  takes  up  all  the  direct  weight  of  the  leaf,  the  upper  hinge  taking  up  only  hori- 
zontal pull.  Each  leaf  is  opened  and  closed  by  means  of  an  operating  strut  AB, 
pinned  to  the  top  of  the  leaf  and  to  the  rim  of  a  horizontal  bull-wheel;  each  wheel 
is  driven  by  an  engine  through  a  train  of  gears.  Each  leaf  weighs  463,000  lbs.;  the 
distance  from  its  center  of  gravity  to  the  (vertical)  axis  of  its  hinges  is  31  ft.  8  ins.; 
the  distance  between  the  hinges  is  48  ft.;  the  diameter  of  the  upper  hinge  pin  is  12 
ins.;  the  radius  of  the  pintle  is  9  ins.  Assume  coefficients  of  friction  for  pin  and 
pintle  to  be  0.05  and  0.15  respectively.  Determine  the  reactions  at  the  hinge  pin 
and  pintle  due  to  the  weight  of  the  leaf,  and  the  moment  of  the  frictional  resistance 
to  swinging,  about  the  axis  of  the  hinges. 

i66-(45).  Fig.  87  represents  certain  details  of  the  operating  mechanism  for  the  lock 
gate  described  in  the  preceding  problem.  It  will  be  noticed  that  when  the  gate  is 
wide  open,  the  axis  of  the  operating  strut  is  over  the  center  of  the  bull-wheel.  The 
dimensions,  proportions,  etc.,  are  such  that  a  turn  of  the  wheel  through  180°  from 
the  position  shown  closes  the  leaf;  and  then  the  center  of  the  wheel  is  again  in  Hne 
with  the  axis  of  the  strut.  From  a  large  drawing,  we  have  scaled  the  arms  of  the 
thrust  of  the  strut  with  respect  to  the  axes  of  rotation  of  the  leaf  and  bull-wheel  for 
thirteen  positions  of  the  strut,  corresponding  respectively  to  the  open  position  of  the 
wheel,  15°  turn,  30°  turn,  etc.  (see  adjoining  table).  Compute  the  torque  required 
on  the  bull-wheel  for  overcoming  the  frictional  resistance  at  the  hinges  for  each  of 
the  thirteen  positions,  neglecting  the  frictional  resistance  at  the  pins  of  the  strut, 
at  the  center  pin  of  the  wheel  and  at  the  rollers  under  the  rim  of  the  wheel.  Make  a 
curve  which  shows  how  this  torque  varies  with  6.  What  docs  the  area  under  the  curve 
represent? 


348 


Position 

e,  degrees 

a',  feet 

a",  feet 

Open     I 

0 

19.9 

0.0 

2 

IS 

18 

8 

1.8 

3 

30 

18 

2 

3.8 

4 

45 

18 

4 

5-8 

5 

60 

19 

2 

7.8 

6 

75 

20 

3 

95 

7 

90 

21 

5 

10.7 

8 

105 

22 

5 

II. 2 

9 

120 

23 

0 

10.7 

lO 

135 

22 

9 

9-4 

II 

150 

22 

3 

7.0 

12 

165 

21 

3 

3-7 

Shut    13 

180 

19 

9 

0.0 

Plan 


Section  at  Bull  Wheel. 


Fig.  87 


349 

i67-(45)-  The  shaft  of  the  engine,  which  operates  the  gate  described  in  the  pre- 
ceding problems,  carries  a  pinion  A  (Fig.  87);  A  drives  a  spur  wheel  5  on  a  second 
shaft  which  also  carries  a  bevel  pinion  C;  C  drives  a  bevel  spur  wheel  Z)  on  a  vertical 
shaft  which  also  carries  a  pinion  E;  E  gears  with  the  rack  F  and  thus  drives  the  bull- 
wheel.  The  numbers  of  teeth  on  the  pinions,  wheels,  etc.,  are  as  follows: 
A  B  C  D  E  F 

18  126  20  90  16  150 

the  last  for  one-half  the  circumference,  more  not  required.  Neglecting  all  friction 
loss  in  the  operating  mechanism,  compute  the  torque  required  at  the  engine  shaft 
for  each  of  the  thirteen  positions  of  the  gate  mentioned  in  the  preceding  example, 
and  make  a  graph  which  shows  how  this  torque  varies  with  e.  Recompute,  but 
allow  for  friction  loss  by  means  of  (estimated)  efficiency  of  the  gear  train.  What  is 
the  total  amount  of  work  done  at  the  engine  shaft  in  closing  one  gate? 

i68-(45).  The  engine  (preceding  problems)  is  run  at  370  rev/min.  Compute 
the  rate  (in  horse-power)  at  which  the  engine  works,  at  the  engine  shaft,  when  closing 
a  leaf  at  each  of  the  thirteen  positions  mentioned.  Show  by  means  of  a  graph  how 
the  power  varies  with  0  and  the  time.  (The  author  is  indebted  to  Mr.  B.  H.  Parsons, 
Mechanical  Engineer  of  the  Mississippi  River  Power  Company,  for  the  data  of  these 
problems  relating  to  the  Keokuk  Lock.) 

i69-(46).  Water  is  flowing  through  a  certain  6-in.  pipe  at  a  velocity  of  4  ft/sec. 
Compute  the  resultant  pressure  of  the  water  against  a  right-angle  bend  in  the  pipe. 
(Assume  that  the  water  pressure  is  the  same  at  both  ends  of  the  bend,  and  equals 
100  lbs/in^.) 

1 70- (46).  Actually,  the  water  pressure  (referring  to  the  preceding  problem)  is 
greater  at  the  inlet  end  of  the  bend.  Assume  that  the  pressures  are  104  and  100 
lbs/in^;  then  solve. 

i7i-(46).  A  certain  three  and  one-half  inch  hose  is  conducting  water  at  a  velocity 
of  20  ft /sec.  There  is  a  circular  bend  of  180°  in  the  hose;  the  radius  of  the  bend  is 
8  ft.  Assume  water  pressure  at  both  ends  of  the  bend  to  be  100  lbs/in^.  Determine 
the  resultant  water  pressure  on  the  bend.  How  much  pressure  (tending  to  straighten 
the  hose)  is  there  per  inch  of  bend. 

i72-(46).  Water  is  projected  into  a  smooth  channel  with  borders  so  that  the  mag- 
nitude of  the  velocity  of  the  stream  is  not  changed,  only  its  direction.  Determine 
the  pressure  of  the  stream  against  the  channel. 

i73_(4y).  A  body  weighing  800  lbs.  is  dragged  along  a  smooth  floor  by  a  hori- 
zontal force  which  varies  uniformly  with  the  displacement,  the  force  being  zero 
when  the  displacement  =  o  and  40  lbs.  when  the  displacement  =  10  ft.  Initial 
velocity  (when  5  =  o)  is  2  ft/sec.  Determine  the  time-average  value  of  the  force  for 
the  10  ft. 

i74-(47).  Fig.  88  is  a  part  copy  of  a  figure  from  a  report  on  certain  tests  of  an 
hydraulic  (railway)  buffer  by  Mr.  Carl  Schwartz,  published  in  the  Journal  of  the 
American  Society  of  Mechanical  Engineers  for  June,  1913.  An  abstract  of  the  report 
is  printed  in  Engineering  News  for  Sept.  11,  1913.  The  buffer  consists  essentially 
of  a  cylinder  22  ins.  in  diameter,  and  a  piston;  the  working  stroke  is  11  ft.  The 
buffer  is  firmly  anchored  at  the  stopping  point,  with  the  piston  rod  in  the  line  of  ap- 
proach of  the  buffer  of  the  car  or  locomotive  to  be  stopped.  The  cylinder  is  grooved 
so  as  to  allow  water  to  pass  by  the  piston  during  a  stop. 

The  curve  marked  "speed"  shows  how  the  speed  of  the  locomotive,  in  this  instance, 
varied  during  the   12  sees,  preceding  impact,  and  also  during  the  impact.     Thus 


350 

the  speed  was  about  5.6  mi/hr  at  the  beginning  of  the  test;  it  increased  to  about 
7.3  in  8^  sees;  then  it  decreased  uniformly  up  to  the  instant  of  impact  after  which  it 
decreased  much  more  rapidly.  The  curve  marked  pressure  shows  how  the  hydrauhc 
pressure  behind  the  piston  varied  during  the  impact.  Thus  the  initial  pressure  on 
each  side  of  the  piston  was  about  45  lbs/in^;  after  the  instant  of  impact  the  pressure 
shot  up  to  a  maximum  of  925  lbs/in'^,  and  then  decreased  to  about  80.  The  entire 
travel  of  the  piston  in  this  case  was  3  ft.  (not  indicated  in  the  figure).  The  locomo- 
tive weighed  100  tons. 


10 


1: 

3 
O 

X 


o.  ,.^- 


u 

0) 
D- 


/ 

'N 

^ 

"*■*  . 

5p/-^ 

>,/'■ 

\ 

1 

— -' 

1 

\ 

\ 

1 
\ 

\ 

• 

'^ 

> 

^ 

Pre 

55urt 

'         , 

?■■""■ 

• 

— ^ 

"^ 

1 

000 


soo 


600 


c 


400     t 


200 


•a 


IZ  &  4  O 

Seconds  before  StriKing. 


0.5  1.0  1.5 

Seconds  after  Striking. 


2.0 


Fig.  88 


Compute  the  time-average  and  the  space-average  force  which  stopped  the  loco- 
motive, neglecting  the  effect  of  the  so-called  train  resistance.  Estimate  the  train 
resistance  from  the  retardation  of  the  locomotive  just  before  the  impact,  and  then 
recompute  the  averages  just  mentioned.  Measure  the  area  under  the  pressure  curve 
and  interpret  it.     Does  the  shape  of  the  curve  suggest  any  improvement  in  the  buffer? 

i75-(48).  The  power  of  an  operating  hydraulic  turbine  equals  the  product  of  the 
angular  velocity  of  the  turbine  and  the  rate  at  which  angular  momentum  (about 
the  axis  of  rotation)  of  the  flowing  water  is  changed  in  its  passage  through  the  tur- 
bine.    Prove. 

i76-(48).  A  certain  homogeneous  prism  is  2  X  6  X  36  ins.  in  dimensions.  It 
is  mounted  so  that  it  can  oscillate  like  a  common  pendulum  about  either  of  two  axes 
of  suspension.  Both  axes  contain  the  center  of  one  small  face  of  the  prism;  one 
axis  is  parallel  to  the  2-in.  edges,  and  the  other  is  parallel  to  the  6-in.  edges.  Locate 
the  center  of  percussion  for  each  of  these  axes. 

i77-(4g).  Describe  the  gyrostatic  reaction  which  a  screw-propelled  ship  sustains 
when  pitching  (in  a  rough  sea). 

i78-(49).  In  the  General  Electric  Review,  Vol.  IX,  pages  117  and  118,  there  appears 
the  following:   "The  spin  of  a  precessing  body  increases  the  centrifugal  force  about 


351 

the  axis  of  precession.  Take  the  case  of  a  wheel  spinning  about  a  horizontal  axis 
supported  at  one  end  which  is  precessing  about  a  vertical  axis  through  the  point  of 
support.     The  total  centrifugal  force  is 

which  equals  the  ordinary  centrifugal  force  WV^/gR  plus  the  additional  centrifugal 
force  due  to  spin  (gyroscopic  centrifugal  force)  {WV^k'^p-)/{gR2r'^).  W  =  weight  of 
the  gyroscope,  k  —  its  radius  of  gyration,  R  =  the  radius  of  the  circle  of  precession, 
r  =  the  radius  of  the  spinning  wheel,  V  =  the  hnear  velocity  of  the  precession,  v  = 
the  peripheral  velocity  of  the  wheel,  and  p  =  the  ratio  v/V.'"  Presumably,  R  means 
the  radius  of  the  circle  described  by  the  mass-center  of  the  wheel.  Ascertain  in  your 
own  way  whether  any  force,  appropriately  called  centrifugal  force,  has  the  value 
above  stated  in  the  case  in  question. 

i79-(49).  On  page  144  of  the  journal  mentioned  in  the  preceding  problem  there 
appears  this  statement.  "The  total  vertical  force  on  the  outside  rail  [car  wheels 
running  around  a  curve]  due  to  gyroscopic  action  will  therefore  be  (3  WV^k"^)  -i- 
{2gRrx)."  W  =  the  weight  of  a  pair  of  wheels  and  axle  (presumably),  k  =  radius 
of  gyration  of  the  pair  and  axle  (about  their  axis) ,  r  =  the  radius  of  the  wheels,  R  = 
radius  of  the  curve,  x  =  gage  of  the  track,  and  V  =  the  velocity  of  the  car.  Can 
you  prove  the  statement? 

i8o-(5o).  A  wheel  6  ft.  in  diameter  rolls  on  a  straight  track.  At  a  certain  instant 
the  velocity  and  acceleration  of  its  center  are  10  ft/sec.  and  4  ft /sec/sec.  Deter- 
mine the  acceleration  of  the  lowest  point  of  the  wheel  at  the  instant  in  question. 

i8i-(5i).  When  a  slender  body,  such  as  a  pole,  chimney,  etc.,  is  tipped  over  from 
an  upright  position,  the  motion  is  one  of  rotation  about  the  point  of  contact  of  the 
body  and  the  surface  which  supports  the  body  until  slip  occurs  at  the  contact  or 
the  lower  end  Ufts  from  the  surface.  Assume  that  the  slender  body  is  hinged  to  the 
supporting  surface  so  that  it  cannot  slip  or  Hft,  and  then  determine  the  vertical  and 
horizontal  components  {V  and  H)  of  the  supporting  force  for  various  positions  of  the 
tipping  body.  Draw  curves  showing  how  V  and  H  vary  with  the  angular  displace- 
ment of  the  pole  from  the  vertical.  How  could  you  ascertain  whether  slip  or  hft 
would  occur  first? 

i82-(5i).  Referring  to  the  preceding  problem,  assume  that  the  pole  is  supported 
on  the  ground,  and  that  sUp  cannot  occur  during  tipping.  The  lower  end  of  the  pole 
will  lift  when  a  certain  degree  of  tip  is  reached;  afterwards  the  pole  moves  under  the 
influence  of  gravity  only.  Until  the  pole  strikes  ground,  it  rotates  with  the  angular 
velocity  which  it  had  at  the  instant  when  the  contact  was  broken,  and  the  center  of 
gravity  moves  in  a  parabolic  path  due  to  its  initial  velocity  (when  the  contact  was 
broken)  and  action  of  gravity.  Determine  the  distance  from  the  (original)  point 
of  support  of  the  pole  to  where  it  first  strikes  the  ground. 

i83-(52).  In  Fig.  410,  the  load  W  =  18,000  lbs.;  the  diameter  of  the  roUers  = 
15  ins.;  the  coefficient  of  rolling  resistance  "under"  the  rollers  =  0.020,  that  "over" 
the  rollers  =  0.025.  How  large  a  force  P  is  required  to  move  the  load?  Determine 
the  two  forces  which  act  upon  a  roller  supposing  that  the  load  is  distributed  equally 
among  the  rollers. 

i84-(52).  Referring  to  Prob.  162:  The  rollers  used  were  3  ins.  in  diameter;  about 
2000  were  used.    They  were  of  steel  2  ft.  long  and  rolled  between  steel  plates  above 


352 

and  below.  Assume  that  yoiir  computed  result  in  Prob.  162  is  the  value  of  the  pull 
actually  exerted  on  the  building  when  moving  on  a  level  stretch.  Then  compute 
the  average  coefficient  of  rolling  resistance. 

i85-(53).  Two  men  A  and  B  are  walking  at  a  speed  of  4  mi/hr  along  east  and  west 
and  north  and  south  paths  respectively.  Compute  the  velocity  of  A  relative  to  B 
when  A  is  walking  northward  and  B  eastward;  when  A  is  walking  northward  and 
B  westward. 

i86-(s3).  The  disk  (Fig.  89)  is  4  ft.  in  diameter  and  is  rotating  uniformly  about  O 
at  one  rev/sec.  A  point  P  is  moving  uniformly  along  the  diameter  AB  from  A 
toward  B  at  a  speed  of  4  ft/sec.  Determine  the  absolute  velocity  of  P  when  midway 
between  A  and  0;  when  midway  between  0  and  B. 

i87-(53).  Suppose  that  P  (see  preceding  problem)  is  moving  from  C  toward  A; 
the  angle  <i>  —  150°,  and  when  P  reaches  A  its  speed  is  6  ft /sec  (along  CA).  What  is 
the  absolute  velocity  of  P  then? 


Fig.  89 


i88-(53).  A  certain  square  is  6  X  6  ft.,  and  its  corners  are  lettered  A,  B,  C,  and  D 
in  succession  around  the  perimeter.  The  square  is  rotating  uniformly  about  a  Une 
through  .4  perpendicular  to  its  plane  at  one  rev /sec;  a  point  P  is  moving  uniformly 
along  CD  and  in  that  direction  at  6  ft/sec.  Determine  the  absolute  velocity  and 
acceleration  of  P  when  it  reaches  the  mid  position  between  C  and  D. 

i89-(54).  The  sphere  (Fig.  90)  is  suspended  from  the  end  of  a  vertical  shaft  OZ 
by  means  of  the  rod  OC  extending  into  and  rigidly  fastened  to  the  sphere.  The 
shaft  and  the  rod  are  connected  by  a  Hooke's  (flexible)  joint.  When  the  shaft  is 
rotated  it  exerts  a  torque  on  the  rod  which  in  turn  makes  the  sphere  roll  around  on 
the  cone.  Assume  that  the  sphere  is  2  ft.  in  diameter,  R  =  4  ft.,  /  =  8  ft.,  and  that 
the  shaft  makes  1 50  rev/min.  Determine  the  angular  velocity  of  the  sphere,  and  the 
X,  y,  and  z  components  of  that  velocity. 

i9o-(55).  Referring  to  the  preceding  problem,  suppose  that  the  sphere  is  cast 
iron  (weighing  450  lbs/ft^).  Then  compute  the  angular  momentimi  of  the  sphere 
and  determine  the  rate  at  which  the  angular  momentum  is  changing. 

i9i-(55).  Suppose  that  there  is  no  "rolling  resistance"  (Art.  52)  between  sphere 
and  cone.  Then  determine  the  following:  normal  pressure  and  friction  between 
cone  and  sphere;  the  torque  which  the  shaft  must  exert  on  the  rod;  and  the  x,  y, 
and  z  components  of  the  supporting  force  at  0. 

i92-(56).  Fig.  91  represents  in  principle  the  Griffin  Mill  for  grinding  cement.  The 
cross  piece  of  the  (upright)  frame  supports  the  upper  (vertical)  shaft  S  by  means 
of  a  thrust  ball  bearing.    The  large  pulley  P  is  rigidly  fastened  to  the  shaft.    The 


353 


pulley  hub  HH  is  extended  downward  and  is  restrained  laterally  by  the  guides  GG, 
thus  virtually  forming  an  extension  of  the  shaft.  The  "roll"  is  rigidly  fastened  to 
the  "roll  shaft"  and  both  are  suspended  on  a  cylindrical  seat  on  the  inside  of  the  hub 
of  the  pulley  as  shown.  Thus  the  roll  and  its  shaft  can  oscillate  like  a  common  pen- 
dulum about  a  perpendicular  to  the  paper  at  0.  K  is  a  cross  head  rigidly  fastened 
to  the  roll  shaft  but  slipping  in  vertical  guides  on 
the  hub  when  the  roU  and  its  shaft  oscUlate  like  a 
common  pendulum.  The  "die"  is  a  hard  metal 
ring  between  which  and  the  roll  the  grinding  of 
the  cement  takes  place  as  explained  presently. 
When  the  mill  is  idle,  the  roU  shaft  hangs  in  a 
vertical  position;  if  the  pulley  be  rotated  the  guides 
in  the  hub  exert  a  torque  on  the  cross  head,  and 
the  roll  shaft  is  made  to  rotate  in  the  vertical 
position  with  the  pulley.  When  it  is  desired  to 
start  the  mill  for  grinding,  the  roll  is  first  puUed 
outward  "with  an  iron  hook,"  and  then  the  power 
is  turned  on  at  the  pulley.  The  roll  shaft  rotates 
with  the  pulley;  promptly,  the  roll  begins  and 
continues  to  roll  on  the  die  (ring),  a  great  pressure 
being  developed  between  roll  and  die.  Material 
to  be  ground  is  fed  into  the  mill  so  that  some  is 
caught  between  the  roll  and  the  die  and  then  pul- 
verized. Suitable  paddles  on  the  lower  side  of  the 
roU  continually  toss  the  material  which  collects  in 
the  recess  of  the  base;  eventually  it  is  caught  be- 
tween roU  and  die. 

It  will  be  noted  that  the  roll  and  its  shaft  constitute  a  large  gyrostat.  We  now 
propose  the  problem  of  determining  the  pressure  between  the  roll  and  the  ring  when 
the  mill  is  operating.  The  makers  (Bradley  Pulverizer  Co.)  state  it  to  be  about  15,000 
lbs.  for  their  giant  size  when  run  at  a  pulley  speed  of  165  to  1 70  rev/min.  The  follow- 
ing data,  approximated  in  some  cases,  was  taken  from  drawings  furnished  by  the 
makers  of  the  mill.  The  die  is  40  ins.  in  diameter  (inside),  8  ins.  high;  from  the  plane 
of  its  top  to  the  point  of  suspension  0  is  5  ft.  4I  ins.  The  roU  weighs  880  lbs.;  its 
larger  diameter  is  24  ins.  The  roll  shaft  weighs  600  lbs. ;  its  length  over  all  is  6  ft. 
9^  ins.;  its  point  of  suspension  0  is  6  ins.  from  the  upper  end;  its  diameter  varies  from 
'  5 1  ins.  at  the  cross  head  to  6|  ins.  at  the  roll  but  the  ends  in  the  cross  head  and  roll 
are  tapered.  For  simpUcity,  make  the  following  approximations:  roll-shaft  uniform 
diameter  is  5I  ins.,  smaller  diameter  of  roll  =  22  ins.,  and  its  thickness  is  8  ins.  As 
a  further  close  approximation  for  locating  center  of  gravity  and  determining  required 
moments  of  inertia,  assume  that  the  roll  is  a  cylinder  23  ins.  in  diameter  and  8  ins. 
thick  (with  5I  ins.  hole  for  the  roll  shaft). 

i93-(57).  A  certain  right  cone  with  a  circular  base  is  homogeneous;  the  diameter 
of  its  base  is  4  ft.;  the  altitude  is  6  ft.;  and  half  the  apex  angle  is  20°.  Determine 
the  radius  of  gyration  of  the  cone  with  respect  to  an  element  of  its  curved  surface. 


>///w////w///////////^/////Y/w/w/////////^' 


Fig.  91 


PROBLEMS;  SET  B 


The  number  in  parentheses  following  a  problem  mimber  refers  lo  the  article  {in  the  body  of  this 
book)  which  pertains  lo  that  problem.  In  many  of  the  problems,  the  data  are  not  fully  given. 
The  missing  ones  are  indicated  in  the  statement  or  figure;  they  are  to  be  supplied. 

2oi-(3).     Fig.  loi  represents  a  wood  frame  to  which  forces  are  appUed  as  indi- 
cated; the  forces  are  applied  by  means  of  ropes  tied  to  nails  in  the  frame.     By  in- 


*- 4 > 

■         1         ' 

'         1         ■ 
1         '         1 

___, — L— r-- 

— ,--j---, — 
•     1     1 

L__^--L_-. 

'         1         1 
1         1         1 

1       1       1 

Fig.  ioi 

spection  of  the  figure,  estimate  where  you  might  drive  the  nail  and  how  you  would 
pull  the  rope  to  get  the  same  effect  as  the  50  and  70  lb.  forces  give.  Determine  com- 
pletely *  the  resultant  of  the  70  and  80  lb.  forces  by  means  of  the  parallelogram  law. 

202-(3).  Determine  completely  the  resultant  of  the  40  and  50  lb.  forces  by 
means  of  the  triangle  law.  (Remember  that  the  arrow-heads  on  the  sides  of  the 
triangle  are  not  confluent,  do  not  point  the  same  way  around.) 

203-(3).  Determine  completely  the  resultant  of  the  60  and  70  lb.  forces  alge- 
braically. 

204-(3).     Determine  completely  the  resultant  of  the  40  and  70  lb.  forces. 

205-(3).  Resolve  fully  the  50  lb.  force  into  two  components  parallel  to  the  60  and 
80  lb.  forces. 

2o6-(3).  Resolve  fully  the  60  lb.  force  into  two  components,  one  perpendicular 
to  that  force  and  one  horizontal. 

207-(3).  Resolve  fully  the  60  lb.  force  into  two  components,  one  perpendicular 
to  that  force  and  one  vertical. 

2o8-(3).  Resolve  fully  the  60  lb.  force  into  two  components,  one  horizontal 
and  one  vertical. 

209-(3).  Resolve  fully  the  60  lb.  force  into  two  components,  one  applied  at  the 
upper  right-hand  corner  and  one  at  the  lower  left-hand  corner  of  the  square. 

*  In  many  of  these  problems  it  is  required  that  one  or  more  forces  be  determined  wholly 
or  in  part.  Complete  determination  requires,  not  only  the  magnitude  of  the  force,  but 
also  its  direction  and  at  least  one  point  in  its  line  of  action;  statement  also  of  magnitude, 
line  of  action,  and  sense  is  complete.  In  order  to  make  the  line  of  action  of  a  required  force 
seem  more  real,  the  student  should  decide  upon  an  appropriate  point  of  application;  for 
example  a  nail  or  hook  in  the  body  to  which  one  could  ti^.  a  cord  or  rope  for  applying  the 
iorce. 

354 


355 

2io-(3).  Resolve  fully  the  60  lb.  force  into  three  components,  applied  along 
the  two  sides  and  bottom  of  the  frame. 

2ii-(3).  Resolve  fully  the  60  lb.  force  into  two  components,  one  of  120  lbs.  applied 
at  point  I  and  one  horizontal. 

2i2-(3).  Is  it  possible  to  apply  three  forces  along  the  sides  of  a  triangular  board 
so  that  they  will  balance?  Is  it  possible  to  apply  four  unequal  forces  along  the  sides 
of  a  square  board  so  that  they  will  balance? 

2i3-(4)-  Compound  completely  the  30,  40,  60,  and  90  lb.  forces  (Fig.  loi)  graphi- 
cally.    (Do  not  draw  the  force  polygon  in  the  space  diagram;  use  standard  notation.) 

2i4-(4).  Compound  completely  the  40,  50,  70,  and  100  lb.  forces  algebraically. 
(Specify  the  direction  of  the  resultant  by  means  of  the  acute  angle  between  its  line 
of  action  and  the  horizontal.) 

2i5-(4).     Resolve  fully  the  100  lb.  force  (Fig.  102)  into  x,  y,  and  z  components. 

2i6-(4).     Compound  completely  the  forces  applied  to  the  cube  (Fig.  102). 


YU---  4   -  ->i 

I 

jo 


Fig.  102 

2i7-(5)-  Compute  directly  (force  times  arm)  the  moment  of  the  50  lb.  force 
(Fig.  loi)  about  point  i. 

2i8-(5).  Compute  indirectly  (by  means  of  the  principle  of  moments)  the 
moment  of  the  60  lb.  force  (Fig.  loi)  about  point  2. 

2i9-(5)-  Without  actually  locating  the  line  of  action  of  the  resultant  of  the 
horizontal  and  vertical  forces  in  Fig.  loi,  ascertain  where  the  line  of  action  cuts  a 
side  of  the  frame.     (Use  principle  of  moments.) 

2  20-(5).  Fig.  103  is  a  cross  section 
of  a  "rolling  dam."  AB  is  the  sheath 
rigidly  fastened  to  the  cylinder  which  can 
be  rolled  upward  on  two  inclined  racks 
CD,  one  at  either  end  of  the  dam.  The 
figure  shows  the  dam  resting  on  the  bed 
at  A  and  against  the  rack  at  B.  In  that 
position  the  horizontal  and  vertical  com- 
ponents, Pi  and  Pi  respectively,  of  the 
water  pressure  are  180  and  30  tons.  The 
weight  W  is  70  tons.  Without  compu- 
ting the  resultant  of  these  three  forces, 


\^vv\\w\\vw 


Fig.  103 


Fig.  104 


ascertain  how  far  from  D  its  line  of  action  intersects  the  Hne  CD. 

22i-(s).  Resolve  each  of  the  forces  shown  in  Fig.  104  into  a  force  applied  at  the 
center  of  the  pulley  and  a  couple. 

222-(5).  The  forces  (Fig.  105)  arc  applied  at  right  angles  to  the  cranks.  Re- 
solve Pi  into  a  force  at  A  and  a  couple,  and  P2  into  a  force  at  B  and  a  couple. 


356 

223-(6).     Fig.  io6  is  a  half-section  of  a  building.     The  four  forces  are  wind  pres- 
sures, perpendicular  to  and  applied  at  mid-points  of  the  portions  oi,  T2,  etc.     Deter- 


R=  100 lbs.  ^ 

1 

n 

B.r" 

1 

.-f 

na 

1 

±- 

7JA' 

\y 

A 

1 

>    W" 

<  ^ 

y^ 

^i- 

100  lbs. 

0 

\/(6,9) 

-X        1 

^  p 

4 

i5 

Fig.  106 

Fig.  105 

mine  completely  the  resultant  of  the  four  forces  graphically.  (Note  where  the 
line  of  action  cuts  'il  and  45.  The  figures  in  parentheses  are  coordinates  of  the 
points  I,  2,  and  3  with  respect  to  O.) 

2  24-(6).     Determine  completely  the  resultant  of  the  three  forces  described  in 
Prob.  220. 

2  2  5-(7).     Determine  completely  the  resultant  of  the  six  forces  applied  to  a  plank 

as  shown  in  Fig.  107. 

226-(7).  What  force  applied  at  the  middle  of  the  plank 
(Fig.  107)  and  couple  are  equivalent  to  the  three  forces 
acting  upward? 

227-(7).     Resolve  fully  the  20  lb.  force  (Fig.  107)  into 
•p       J  two    components   applied   at   A  and   B\   applied    at   B 

and  C. 
Determine   completely  the   resultant   of   the   four  forces   described   in 


A 

B 

> 

C 

<- 

^ 

< 

4- 

•■> 



1 

1 

'Y' 

1 

1 

-+- 

1 

_)_ 
1 



— 

1 

I 

-+- 

_1_ 
1 

-- 

- 

1 

1. 

1 

_J_ 

1 

.  _J_ 

1 
__L_ 

- 

.4'....-— H 


228-(7). 

Fig.  108. 

2  29-(8).  Compute  the  moments  of  each 
force  (Fig.  102)  about  the  coordinate  axes. 

230-(8).     Fig.  109  represents  a  cube  with 
several  forces  applied  to  it.     Specify  as  com- 
pletely as  possible  the  resultant  of  the  two 
couples. 

23i-(8).  Resolve  the  couple  represented 
in  Fig.  105  into  two  components,  one 
whose  plane  is  perpendicular  to  AB,  and  one  whose  plane  is  parallel  to  AB. 

232-(9).     Determine  completely  the  resultant  of  the  forces  excluding  the  two 


Fig, 


Fig.  109 
100  lb.  forces  applied  to  the  cube  shown  in  Fig.  109.     (Specify  the  coordinates  of 
the  point  where  the  line  of  action  of  the  resultant  pierces  the  XY  plane.) 


357 


no 


233-(q)-  Fig.  no  represents  a  three  throw  crank  shaft,  consecutive  cranks  being 
120°  apart.  When  the  shaft  is  made  to  rotate  rapidly,  as  in  a  balancing  machine, 
the  cranks  are  subjected  to  equal  air  pressures.  Assume 
that  these  are  perpendicular  to  the  .cranks  respectively  and 
have  (equal)  4  in.  arms;  then  reduce  these  three  forces  to  a 
force  applied  at  B  and  two  couples. 

234-(io).  (a)  Two  couples  are  in  equiUbrium;  what  can 
you  say  about  them?  (h)  Can  a  force  and  a  couple  be 
in  equilibrium?  (c)  Three  parallel  forces  are  in  equilib- 
rium; what  can  you  say  about  the  middle  force?  {d)  Is 
it  possible  for  four  forces  to  be  in  equilibrium  if  three  are 
parallel?  (c)  Four  forces  are  in  equilibrium  and  two  are 
known  to  constitute  a  couple;  what  can  you  say  about  the  other  two?  (/)  What 
can  you  say  about  the  resultant  of  a  set  of  coplanar  nonconcurrent  forces  whose 
force  polygon  closes? 

surfaces  are  smooth;    a,  /3,  and  7   are  respectively 

,    and  deg.;(2  = 

1000   lbs.      Determine  the   magnitude   of  P 
required  for  equilibrium. 

236-(ii).     The  sphere  (Fig.  112)  weighs  200 
lbs.;    a,  /8,  and  7  are  respectively  , 

,  and  deg.;  all  contact  sur- 

faces are  smooth.     Determine  the  pressures 
on  the  wedge  M  and  the  tension  in  the  rope. 
237-(ii).     A  homogeneous  beam  20ft.  long 
weighs  50  lbs.     At  the  upper   end   it  rests 
held  at  the  lower  end  by  a  rope  ft. 


Ill 


Find  the  angle  between  the  beam  and  the 


against  a  smooth  vertical  wall  and  is 
long  which  is  also  attached  to  the  wall 
wall,  and  determine  the  pull  in  the  rope. 

238-(ii).     Two  right  circular  cylinders  are  supported  in  a  box  18  ins.  wide  as 


shown  in  Fig.  113.     A  weighs 
smooth.      Find    all   of    the 
forces  acting  on  each  cylin- 
der  and   properly  represent 
on  separate  sketches. 

239-(ii).  A  cylindrical 
cask  is  rolled  slowly  along  a 
level  floor  by  means  of  a 
push  applied  at  the  top.  If 
d  is  the  diameter,  W  the 
weight  of  the  cask  and  0  is 


and  B  weighs 


lbs.    All  surfaces  are 


17777777777777777777777777^^ 

Fig.  113 


Fig. 


114 


the  inclination  of  the  force  with  the  horizontal,  determine  the  least  push  required  to 
move  the  cask  over  a  cleat  h  units  high. 

24o-(ii).  AB  (Fig.  114)  is  a  rigid  beam;  two  hooks  are  pinned  to  it  at  A  and  B 
as  shown;  CD  and  CE  are  rods  pinned  to  the  hooks  and  to  each  other;  the  hooks 
engage  a  heavy  body  W.  AB  =  14  ft.,  CD  =  CE  =  8  ft.,  W  =  4  tons.  Deter- 
mine the  tension  in  each  rod  and  all  forces  acting  on  one  hook.  (Neglect  weight  of 
parts.) 


358 


t' 


24i-(ii).     In  Fig.  IIS  ^5  =4  ft.,  AC  =  CD  =  8  ft.;  P  =  200  lbs.;  and  the  cylinder 
E  weighs  100  lbs.     A,  B,  and  E  are  pin  joints.     The  surfaces  at  D  and  F  are  smooth. 

Determine  the  forces  acting  on  the  cyl- 
inder and  those  acting  on  the  bell-crank 
ACD. 

242-(ii).  Fig.  116  shows  Ridley's 
apparatus  for  withdrawing  the  mold 
from  a  freshly  cast  concrete  pile.  The 
device  also  compresses  the  concrete  into 
the  space  vacated  by  the  mold.  Pin 
joints  are  used  at  A,  B,  C,  and  D. 
Assume  P  =  10,000  lbs.;  AB  =  30  ins.; 
BC  =  18  ins.;  and  CC  =  BD  =  24  ins. 
Find  the  force  at  C  and  the  push  on 
the  pile. 


c^ 

; 

^C 

Earth  ' 

:  .■'.■■  >■ 

iiiLil 

^l^/r>-v^ 

riG.  115 


Fig.  116 


-•>k-3'->l 


Fig.  117 


243-(ii).     The  links  in  Fig.  117  are  pinned  at  A,B,  and  C.     A  and  C  are  rigid 
supports.    Assume  P  =  100  lbs.    Find  the  reactions 
(on  the  links)  at  A  and  C. 

244-(i2).  A  horizontal  beam  20  ft.  long  is  sup- 
ported by  a  knife-edge  ft.  from  the  left  end 
and  is  pinned  at  ft.  from  the  right  end. 
It  sustains  loads  of  5000  lbs.  at  the  left  end,  1000 
lbs.  at  10  ft.  from  the  left  end,  2000  lbs.  at  the 
right  end,  and  a  uniformly  distributed  load  of  1000  lbs.  per  ft.  between  the  left  end 
and  left  support.     Determine  the  reactions  of  the  pin  and  knife-edge. 

245-(i2).     The  diagrammatic  sketch  in  Fig.  118  represents  a  lever  system  for  a 
scales.     Assume  W  =  250  lbs.;    a  =  12  ins.;    5  =  8  ins.;    c  =  g  =  ins.; 

(i  =  /z  =  10  Ins.;  e  =  i\  ins.;  /  =  15  ins.     Determine  P. 

r<--a->t< b---;H 


{ 


B 


e—c-fd  ■ 


-*-     I 

I       i 
-*kc->k- 


.d— >i<— d--->k-c->i 


.      !  I 

k--e>W — f ^g>l<—-~ h -»l 

Fig.  118 


- ->k— 5'->S 

Fig.  119 


246-(i3).     The  frame  shown  in  Fig.  119  is  pinned  at  B  and  rests  against  a  smooth 
^      .  ,  .         wall  at  ^.     P  =  1000  lbs.  and  Q  =  250  lbs.     Find  the 

•^1  p  reactions  at  A  and  B. 

247-(i3).     The  beam  in  Fig.  120  weighs  50  lbs.  per 
ft.  and  is  14  ft.  long.     S  is  a  smooth  support;  .4   is  a 

^___^ ^  JF  =  loolbs.  perft.;  P  = 

'<"4-**  ,^^^  iKo  .   rn  _  K-TT  -  K^  ft  •    AF  =  ■<  h      Determine 


Fig.  120 


"<#^   pin  joint.    Q  =  lbs. 

1000  lbs.;  C£>  =  £F  =  6  ft.;  AF  =  i  ft. 
the  reactions  at  A  and  B. 


248-(i3).     A  light  bar  9  ft.  long  is  fastened  to  the  floor  at  its  lower  end  by  a  hori- 
zontal pin  perpendicular  to  the  bar.     The  angle  between  the  bar  and  the  floor  is 


359 


100  lbs. 


deg.;  its  upper  end  rests  against  a  smooth  vertical  wall.  Loads  of  300  and 
200  lbs.  are  hung  at  distances  of  4  and  7  ft.  respectively  from  the  lower  end. 
Calculate  the  reactions  at  the  ends  of  the  bar. 

24g-(i3).     The  uniform  bar  AB  (Fig.  121)  is  pinned  to  the  floor  at  A  and  sup- 
ports a  load  of  100  lbs.  at  5.     The  diameter  of  the  smooth 
cylinder  supporting  the   bar   is  ft.     Determine 

the  forces  on  the  bar,  in  the  rope  AC,  and  on  the   cyl- 
inder. 

250-(i3).     The  truss  shown  in  Fig.  122  is  pinned  at  .1 
and  supported  by  a  smooth  roller  at  B.      Q  =  5000  lbs.; 
W  =  20,000  lbs.;   the  rolling  friction  (along  the  track)  = 
500  lbs.      Compute  the  reactions  at  A  and  B  when  the  car  is   stationary  in   the 
middle  of  the  truss.      (Mass.  Civil  Service  examination.) 

25i-(i3).     The  travelling  wall  crane  (Fig.  123)  rolls  on  three  rails  A,  B,  and  C. 


Fig.  123 

The  weight  of  the  crane  is  i  ton;  the  load  P  is  3  tons;  the  center  of  gravity  of  the 
crane  is  7  ft.  from  the  vertical  through  A.  Neglect  flange  pressures,  and  determine 
the  reactions  on  the  rails. 

252-(i3).  In  a  wall  there  are  two  hori- 
zontal pegs,  A  and  B;  A  is  3  ft.  above  the 
floor,  B  is  6  ft.,  and  the  horizontal  dis- 
tance between  them  is  ft.  A 
straight  bar  15  ft.  long  weighing  200  lbs. 
is  placed  with  its  lower  end  on  the  floor, 
touching  the  underside  of  A  and  the 
upper  side  of  B.  The  bar  is  not  sprung 
into  place,  and  all  surfaces  are  smooth. 
A  weight  of  100  lbs.  is  hung  from  the 
upper  end  of  the  bar.  Determine  the 
pressure  on  the  floor,  on  A ,  and  on  B. 

253-(i3).  A  frame  for  building  and 
raising  a  concrete  wall  is  sketched  in  Fig. 
124.  The  truss  B  is  supy^orted  on  a 
trunnion  D  and  by  the  telescoping  piston 
C  of  a  pneumatic  jack  F.  The  dotted 
lines  show  the  wall  and  jack  in  an  ex- 
treme position.  D  is  10  ft.  vertically  and  12  ft.  horizontally  from  E;  it  is  6  inches 
from  the  upper  surface  IH  of  the  truss;  C  is  3  ft.  from  that  upper  surface;  the 


Fig. 


124 


360 


-b 


^ 


f 
HCK a 


0^ 


e 


■H 


projections  of  C  and  D  on  IH  are  13  ft.  apart.  The  truss  weighs  2  tons;  its  center 
of  gravity  is  i  ft.  below  IH  and  8  ft.  to  the  right  of  D.  The  load  (wall  shown) 
weighs  tons;  its  center  of  gravity  is  ins.  above  IH  and 

ft.  to  right  of  D.     When  the  truss  is  inclined  to  the  horizontal  deg.  as  shown, 

how  large  are  the  pressures  at  C  and  D?     (Solve  graphically.) 

2  54-(i3).     Fig.  125  illustrates  an 

'A  apparatus  for  testing  the  strength  of 

jl3J-^  i      ,^fe?     *^^    small    concrete    beams.      The    force 

produced  by  pouring  shot  into  the 
bucket  L  is  multiplied  by  the  lever 
system  and  imposed  (in  equal  amounts) 
on  the  specimen  B  at  PP,  points 
equally  distant  from  the  center  of  the 
beam.  ^  is  a  knife-edge  penetrating 
the  beam  slightly  and  furnishes  a 
support  for  the  upper  two  levers;  the 
Determine  the  ratio  of  P  to  L. 
126  carries  a  pulley  at  B  weighing  150  lbs. 


a 


"W 


77777777777777m7777m77, 


77777) 


Fig.  125 


other  bearings  on  the  beam  are  rollers. 

255-(i4).  The  vertical  shaft  in  Fig 
and  one  at  E  weighing  80  lbs.  The  radius  of  pulley  B  is  20  ins.,  of  pulley  E  15  ins. 
The  center  of  gravity  i  of  S  is  2  ins.  from  the  axis  of  the  shaft;  f  of  £  is  J  in.  Fi  =  200 
lbs.;  Fi  =  50  lbs.;  and  W  =  200  lbs.  Find  the  reactions  on  the  bearings  at  A  and 
0  when  the  shaft  is  in  the  position  shown. 

256-(i5).     The  truss  represented  in  Fig.  127  is  supported  by  pins  at  .-1  and  B. 
Each  load  P  =  2000  lbs.     Determine  the  amount  and  kind  of  stress  in  each  member. 

257-(i5).    The  truss  represented  in  Fig.  128  is  supported  by  a  pin  at  E  and  by  a 


.'^- 

y 

P 

\ 

-/2.'--> 

<--/2'-4 

Determine 
Each  load 


Fig.  127 

horizontal  tie  at  .4.     Each  load  P\  =  1000  lbs.;  each  loadP2  =  500  lbs 
the  amount  and  kind  of  stress  in  each  member. 

258-(is).     The  truss  represented  in  Fig.  129  is  supported  as  shown 
Pi  =  500  lbs.;  each  load  Pi  =  100  lbs. 
Determine   the   amount   and   kind    of 
stress  in  AB,  AG,  and  BC. 

259-(i5).  The  structure  represented 
in  Fig.  130  is  a  steel  head  frame  for 
hoisting  ore  from  a  mine.  The  frame 
is  pinned  at  A  and  is  anchored  at  B 
so  that  either  an  upward  or  a  down- 
ward reaction  can  occur  at  that  point. 
The  load  is  10  tons.  Determine  the  amount  and  kind  of  stress  in  each  member  of 
the  frame. 

26o-(i5).     The  truss  represented  in  Fig.  131  rests  on  a  smooth  support  at  A  and 


Fig.  129 


361 


is  held  by  a  pin  at  E.    Each  load  Pi  =  500  lbs.;    each  load  Pi  =  1000  lbs.;    each 
load  Pz  =  2000  lbs.     Determine  the  amount  and  kind  of  stress  in  each  member. 
26i-(i5).     The  truss  represented  in  Fig.  132  is  held  by  a  pin  at  joint  B  and  by  a 


Fig.  130 


Hf 40'- ^ 

Fig.  131 

horizontal  tie  at  joint  A.  Each  load  =  1000  lbs.  Solve  for  the  stress  in  each 
member. 

262-(i6).  The  truss  represented  in  Fig.  133  is  supported  at  each  end.  Each 
load  Px  =  1000  lbs.,  and  P2  =  2000  lbs.  Determine  the  amount  and  kind  of  stress 
in  each  member  graphically. 

263-(i6).    The  truss  represented  in  Fig.  134  is  held  by  a  pin  at  the  right  end  and 


->i<--    --^--    —A<—     -->!  A.k--^ — 

Fig.  133  Fig.  134 

rests  upon  a  smooth  support  at  the  left  end.     Each  load  Pi  =  1000  lbs.;  Pi  =  2000 
lbs.     Solve  graphically  for  the  stress  in  each  member. 

264-(i7).  The  crane  represented  in  Fig.  135  is  pinned  at  B  and  is  held  by  the 
guy  rope  CA.  The  load  W  is  20  tons.  PG  =  3  ft.  Determine  the  tension  in  the 
guy  rope  and  all  forces  acting  on  each  member  of  the  crane. 

S'     \<-/0'^.'yP--/z' >K-<9->i 


Fig.  13s 


T 

w 


Fig.  136 


26s-(i7).    The  crane  represented  in  Fig.  136  rests  in  a  socket  at  A  and  bears 
against  the  smooth  side  of  the  hole  in  the  floor  at  D.    There  are  pins  at  B,  C,  and  P. 


362 


B 


The  load  W  is  4000  lbs.,  the  counterweight  E  weighs  5000  lbs.    Determine  all  the 
forces  which  act  upon  each  member  of  the  crane. 

266-(i7).     The  crane  represented  in  Fig.  137  rests 

in  a  socket  at  A  and  passes  through  an  opening  in 

the  floor  at  B.     The  post  AB  passes  freely  through  a 

smooth  slot  in  the  boom  at  C,  so  that  any  reaction 

existing    there    will    be 

horizontal.      The    load 

PF  is  3  tons  and  is  6  ft. 

out  from  C.    Determine 

all  the  forces  which  act 

upon   each  member  of 

the  crane. 

267-(i7).  The  crane 
represented  in  Fig.  138  rests  in  a  socket  at  A  and  is 
horizontally  restrained  by  the  ceiling  at  B.  The  load 
PF  is  4  tons.  The  weights  of  the  members  are:  post 
AB  =  0.7  ton,  brace  DE  =  0.5  ton,  boom  CF  =  0.6  ton.  The  center  of  gravity 
of  the  boom  is  9  ft.  to  the  right  of  C.  For  the  other  members  the  position  of 
the  center  of  gravity  is  midway  between  the  ends.  Determine  all  the  forces  which 
act  upon  each  member. 


Fig.  137 


WW77' 


Fig.  138 


•'jmiiii. 


->K--6-^ 


^•W/WWrr. 


Fig. 


139 


268-(i7).  The  crane 
pinned  at  B  and  is  sup- 
ported horizontally  at  A. 
The  load  IF  is  i  ton.  De- 
termine the  amount  and 
kind  of  stress  in  each  of 
the  members  C£,  DE, 
and  BE,  and  determine 
all  forces  acting  upon  the 
post  and  boom. 

26g-(i8).      The  crane 


represented    in    Fig.    139  is 


140 


shown  in  Fig.  140  rests  in  a  socket  at  A  and  passes 

through  a  hole  in  the  floor  at  B,  the  sides  of  the  hole  affording  horizontal  support. 


r  ■-     -f^^ ^ 


Fig.  141 

The  diameter  of  each  pulley  is  2  ft 
to   pulley  on    the   post  at  B. 
on  each  member. 


Fig.  142 

The  rope  is  vertical  from  F  to  D  and  is  fastened 
The  load  W  is  8  tons.    Determine  all  forces  acting 


363 


to; 


Fig.  143 


27o-(i8).  In  Fig.  141  the  members  are  pinned  to  the  wall  at  A  and  at  i5,  and  to 
each  other  at  C.  The  diameter  of  the  pulley  is  2  ft.;  the  load  W  is  1000 lbs.  Deter- 
mine the  forces  which  act  upon  each  member  of  the  structure. 

27i-(i8).  The  crane  shown  in  Fig.  142  is  pinned  at  A  and  is  horizontally  re- 
strained by  the  floor  at  B.  The  diameter  of  the  pulley  at  Z)  is  2  ft.,  of  the  drum  on 
the  floor,  i  ft.  The  load  W  is  2  tons.  Determine  all  the  forces  which  act  upon 
each-  member. 

272-(i9).  A,  B,  and  C  (Fig.  143)  "respectively  weigh  100,  200,  and  300  lbs.  The 
surfaces  in  contact  are  so  rough  that  a  force  P  of  lbs. 

causes  no  slip.     Represent  in  a  separate  sketch  of  each  body  all 
the  forces  acting  on  it  when  P  has  the  stated  value. 

273-(i9).  A  body  is  on  a  horizontal  surface  and  is  subjected 
to  a  force  which  does  not  make  it  move;  the  body  weighs  200  lbs.; 
the  force  is  incUned  deg.  with  the  horizontal  and  equals 

200  lbs.     Determine  the  normal  pressure  and  friction  on  the  body  when  the  force 
is  a  pull;  when  a  push. 

2  74-(i9).     Suppose  that  the  coefficient  of  friction  for  A   and  B  (Fig.  143)  is 
;   for  B  and  C  ;   and  for  C  and  D  .    What  is  the  least 

value  of  P  which  would  cause  a  slip? 

275-(i9).     The  coefficient  of  friction  between  A   and  B  (Fig.   132,  Art.   19)  is 
and  e  =30°.      Determine   the   least   pull  P  and   push  P  which  would 
cause  slip. 

276-(i9).  Compare  the  values  of  the  pull  P  required  to  cause  slip  (Fig.  132, 
Art.  19)  when  0  =  o,  e  =  30,  and  9  =  50°. 

277-(i9)-  A  straight  bar  rests  in  a  vertical  plane  with  one  end  on  a  rough  hori- 
zontal floor  and  the  other  against  a  smooth  vertical  wall.  The  coefficient  of  fric- 
tion for  floor  and  bar  is  .  At  what  minimum  angle  between  bar  and  floor 
would  the  bar  rest? 

278-(i9).  Suppose  that  the  bar  of  the  preceding  problem  weighs  100  lbs.,  and  is 
set  at  an  angle  of  deg.    Determine  the  necessary  downward  force  applied 

at  the  upper  end  to  cause  slip  of  the  bar. 

279-(i9).  The  ladder  AB  (Fig.  144)  is  40  ft.  long  and  weighs  100  lbs.  The  co- 
efficient of  friction  at  A  (between  the  upper  rung  and  the  pole)  is  ^;  at  5  j.  Com- 
pute the  force  P  required  to  overcome  gravity  and  friction  in  the  position  shown. 

-ffod 

'to 


Fig. 


Fig. 


144  riG.   145 

28o-(i9).  Fig.  145  represents  the  cross  section  of  a  dam,  a  sluice  gate  G,  and  a 
log  sluice  or  trough  AB  (shown  in  section  at  S).  Water  is  shown  passing  over  the 
gate  and  down  the  sluice  permitting  the  passage  of  logs.  The  sluice  is  made  ad- 
justable to  the  water  level.    The  front  wheels  at  A  rest  against  vertical  rails,  and 


364 


the  wheels  at  B  on  indined  rails.  The  end  A  is  raised  or  lowered  by  means  of  a 
vertical  rod  operated  from  above  by  a  suitable  winze.  The  log  sluice  weighs  10  tons, 
AB  =  80  ft.  Determine  the  pull  at  the  rod  required  to  overcome  the  weight  of 
the  sluice  and  the  friction  at  A  and  B  in  the  position  shown.  (The  diameter  of  the 
wheels  A  and  B  is  small  compared  to  AB;  so  regard  the  sluice  as  slipping  on  the 
two  ends  (like  the  ladder  in  the  preceding  problem)  and  take  the  equivalent  coeffi- 
cient of  friction  as  tV-) 


28i-(2o).    A  (Fig.  146)  weighs  100  lbs. 


30' 


when  /3  =  o  and  P  =  lbs.,  A 

does  not  move,  (a)  Compute  the  fric- 
tion on  A  and  determine  its  direction. 
(b)  Solve  when  /?  =  40°. 

282-(2o).  A  (Fig.  147)  weighs  100 
lbs. ;  the  coefficient  of  friction  under  A 
is  J.  What  force  P  is  required  to  pull 
.1  down? 

283-(2o).     The  coefficients  of  friction 
between  A  and  B  and  B  and  C   (Fig. 
148)  are  \.     W  weighs  10  tons.      How 
great  must  P  be  to  start   the  wedge  B? 

284-(2o).     A  (Fig.  149)  weighs  10,000  lbs.;   the  coefficient  of  friction  for  all  con- 
tacts is  ^.     What  value  of  P  is  required  for  starting  A  up  the  plane? 


Figs.  146,  147 


St 


BL 


Fig.  149  Fig.  150 

28s-(2o).  In  Fig.  150  W  =  7000  lbs.;  the  coefficient  of  friction  for  all  contacts 
is  \.  Determine  the  values  of  horizontal  forces  applied  to  wedges  B  required  for 
raising  W . 

286-(2o).  The  required  forces  mentioned  in  the  preceding  problem  are  supplied 
by  means  of  the  right-and-left  screw,  Fig.  1 50.  The  mean  diameter  of  screw  threads 
is  ins.,  and  the  number  of  threads  per  inch  is  Determine  the 

torque  required  for  raising  W . 


Fig.  151 

287-(2i),  Each  parallel opiped  {A,  B,  C,  Fig.  151)  is  homogeneous;  their  weights 
are  respectively  8,  42,  and  16  lbs.  Determine  the  distances  of  their  center  of  gravity 
from  the  coordinate  planes. 

288-(2i).  Fig.  152  represents  a  bent  wire.  Determine  the  coordinates  of  its 
center  of  gravity. 


365 


289-(2i).  Fig.  153  represents  a  cube  with  two  grooves  cut  along  the  medians  of 
two  sides  as  shown.  The  grooves  are  2  ins.  wide  and  i  in.  deep.  Determine  the 
coordinates  of  the  center  of  gravity  of  the  grooved  cube. 


t<-  4">t  i<  4">t 


Ok- 8"- -MX 

Fig.  154 


_x 


29a-(22).     Determine  the  coordinates  of  the  centroid  of  the  shaded  area  in  Fig. 
154.     (See  Art.  24  for  centroid  of  semicircular  area.) 

29i-(22).  The  angles  Fig.  155  are  6X4X5.  Locate  the  centroid  of  the  entire 
shaded  area. 

292-(2  2).     Determine  the  centroid  of  the  shaded  area  in  Fig.  156.    The  area  of  the 
hole  is  8  sq.  ins.  and  the  coordinates  of  its  centroid   are  x  =  2> 
and  y  =  ^  ins. 

293-(23).  Imagine  an  ellipsoid  separated  into  halves  by  one 
of  its  planes  of  symmetry.  Determine  the  position  of  the  cen- 
troid of  one  of  these  halves. 

294-(24).  Prove  the  statements  on  page  100  about  the  cen- 
troid of  a  right  circular  cylinder. 

29S~(25)-     A  certain  wire  weighs  lbs/ft,  and  can  sustain 

a  pull  of  lbs.  with  safety.     It  is  to  be  suspended  between  two  points  on  the 

same  level  and  1000  ft.  apart.     Assume  that  the  suspended  wire  will  be  parabolic, 
and  compute  the  shortest  piece  of  wire  that  may  be  used. 

296-(26).  Solve  the  preceding  problem  but  on  the  assumption  that  the  sus- 
pended wire  takes  the  catenary  form. 

297-(27).  A  rope  is  to  be  suspended  at  its  two  ends  from  two  points  AB  on  the 
same  level  and  40  ft.  apart.  Heavy  weights  are  to  be  hung  from  knots  on  the  rope 
so  that  the  rope  will  assume  the  form  of  half  an  octagon.  What  weights  will  hold 
the  rope  to  the  desired  form? 


298-(28).  The  following  data  were  secured  from  the  launching  of  the  battleship 
Conneclicut  {Engineering  News  for  Dec.  22,  1904): 

Time  /,  5        10      15       20       25      30      35      40      45       50    sec. 

Distances,        18        60     132     225     325    420    520    610    680    720    ft. 
Determine  the  velocity  when  t  =  sec;  when  /  =  sec. 

299-(28).  A  point  moves  in  a  straight  line  in  accordance  with  the  law  s  =  ^  —  40  t, 
where  s  is  distance  in  feet  from  a  given  point  in  the  path  and  t  i?  time  in  seconds.  By 
calculus  find  the  velocity  when  t  =  sec.     What  is  the  average  velocity  for  the 

second  preceding  the  instant  named?  for  the  second  following  the  instant? 


366 

3oo-(28).  Express  a  velocity  of  60  mi/hr  in  feet  per  second.  A  certain  electric  car 
gets  up  such  a  velocity  in  sees.     Express  that  average  starting  acceleration  in 

foot  and  second  unit;  compare  with  "gravity." 

30i-(28).  In  a  certain  run  the  velocity  of  an  electric  car  changed  in  the  following 
manner: 

Timet,  o     10     15     20        25        30        35        40     50        114     125  sec. 

Velocity^),       o     15     21     24.5     26.5     28.5     29.5     31     32.5       21        o  mi/hr. 

The  car  coasted  during  the  interval  50-114  and  was  braked  during  the  interval  114- 
125.  (Sheldon  and  Hausman,  Electric  Traction  and  Transmission  Engineering,  p.  63.) 
Find  the  acceleration  when  t  —  sec;  when  /  =  sec. 

302-(28).  In  a  certain  rectilinear  motion  v  =  /-  —  10  /,  where  v  is  velocity  in  feet 
per  minute  and  /  is  time  in  minutes.  Determine  by  calculus  the  acceleration  when 
t  =  min.     What  is  the  average  acceleration  for  the  minute  preceding  the  instant 

named?  for  the  minute  following  the  instant? 

303- (28).  A  point  moves  in  a  straight  line  in  accordance  with  the  law  5  =  2  sin 
(0.05  /  +  ).  where  5  is  in  inches,  /  in  seconds,  and  the  angle  in  radians.     Deter- 

mine the  velocity  and  acceleration  when  t  =         ;  when  /  =  .     Interpret  the 

signs  of  your  results. 

304-(28),  A  body  has  an  "initial  velocity"  (when  /  =  o)  of  15  ft /sec  and  an 
acceleration  expressed  by  a  =  90  /  —  24  /-,  the  units  being  foot  and  second,  (a) 
How  far  does  the  body  move  in  the  interval  /  =  to  /  =  ?     (b)  What  are  the 

velocities  at  the  instants  named?     (c)  What  is  the  average  velocity  for  the  interval? 

305-(28).  In  a  certain  series  of  tests  on  emergency  stops  with  an  automobile  the 
following  data  were  obtained: 

Speed  of  auto,  6  10  15  20  25  30         mi/hr. 

Stopping  distance,  1.67         6.00  9.50      36.8         42.0        47.5     ft. 

{Engineering  News  for  Sept.  7  and  Oct.  10,  191 2.)     Compare  the  average  retardations. 

3o6-(29).  Draw  the  distance-time  curve  for  the  data  in  Prob.  298  and  deter- 
mine the  velocity  when  /  =  sec;  when /=  sec.  What  can  you  state  con- 
cerning the  acceleration  at  /  =  30  sec? 

307-(29).  Draw  the  velocity-time  graph  for  the  data  in  Prob.  301.  Determine 
the  acceleration  when  /  =  sec.     What  is  distance  covered  from  /  =  to 

/  =  sec? 

3o8-(29).  A  train  can  get  up  a  speed  of  60  mi/hr  in  5  min.,  and  stop  in  0.5  mi. 
About  midway  between  two  stations  10  mi.  apart  a  bad  piece  of  track  one  mile  long 
necessitates  reduction  of  speed  to  10  mi/hr.  Assuming  that  acceleration  and  re- 
tardation can  be  applied  uniformly  with  respect  to  time,  determine  the  time  between 
stations.     (Sketch  the  velocity-time  graph  before  calculating.) 

309-(29).  In  the  preceding  problem  how  much  time  was  lost  on  account  of  de- 
fective track? 

3io-(29).  In  testing  automatic  safety  cushions  placed  at  the  bottom  of  elevator 
shafts  in  the  Wool  worth  Building,  N.  Y.,  the  following  data  were  obtained: 

Distance   from   top 

of  air  cushion,  o       20      40       60       80     100     120     130     135     137  ft. 

Pressure  on  bottom 

of  car,  4        4        7, 10      12        9        9        9        8        o  Ib/in^, 

Downward  velocity 

of  elevator,  168     168     157     140    116      92      67      52      32        oft/sec. 


367 

Plot  the  distance-velocity  curve,  and  find  the  acceleration  when  the  elevator  had 
fallen  ft.  in  the  cushion.     {Engineering  Record  for  Sept.  5,  1914-) 

31 1-(29).  From  the  data  in  the  preceding  problem  find  the  time  consumed  in  fall- 
ing ft.  in  the  air  cushion.  (Hint.  Plot  the  reciprocals  of  the  velocities  against 
the  corresponding  distances.) 

3ii-(3i).  The  following  quotation  from  Engineering  News,  Dec.  3,  1914,  relates  to 
tall  building  elevators:  "I  and  some  of  my  associates  have  been  subjected  to  retarda- 
tions as  high  as  72  ft /sec/sec  without  much  discomfort.  The  stress  in  the  ankles  is 
quite  noticeable.  In  air  cushion  practice  it  is  customary  to  allow  for  retardations  of 
five  or  even  six  times  that  of  gravity;  i.e.,  retardations  up  to  nearly  200  ft /sec /sec, 
and  it  is  considered  that  even  this  high  retardation  will  not  be  injurious  to  hfe.  There 
are  several  instances  on  record  where  it  has  been  sustained  without  serious  injury." 
Calculate  in  the  case  of  a  man  standing  upright  in  an  elevator  undergoing  a  retarda- 
tion of  six  times  gravity,  the  pressure  on  the  soles  of  his  feet,  and  the  stress  at  his  neck. 
(Take  the  weight  of  his  head  as  7  per  cent  of  his  total  weight.) 

3i2-(3o).  The  period  of  a  certain  simple  harmonic  motion  is  '  sec,  and  its 
amplitude  is  ins.     Compute  the  maximum  velocity,  the  velocity  at  a  quarter 

point,  and  the  average  velocity  for  one-half  the  length  of  path.     Ditto  for  acceleration. 

3i2-(3i).  Take  the  weight  of  the  elevator  in  Prob.  310  as  7500  lbs.,  and  the  area 
of  its  bottom  as  30  ft'.  Determine  the  amount  and  direction  of  the  resultant  of  all 
forces  acting  on  the  elevator  when  it  was  ft.  below  the  top  of  the  air  cushion. 

What  forces  make  up  the  resultant?     Are  the  records  consistent? 

3i3-(3i).  Imagine  the  surface  C  (Fig.  58)  tilted  so  that  its  inclination  to  the 
horizontal  is  20  deg.;  that  the  coefficient  of  static  friction  between  A  and  B  is  0.5; 
that  the  coefficient  of  kinetic  friction  between  B  and  the  incline  is  0.2.  A  and  B 
weigh  100  and  lbs   respectively,  and  P  =  lbs.     Find  the  acceleration  of 

the  system,  and  represent  the  forces  acting  on  each  body  in  separate  sketches. 

3i4-(3i).  In  the  preceding  problem  determine  the  force  P  which  would  make  A 
slip  on  B. 

3i5-(3i).  Bodies  .4  and  B,  Fig.  157,  weigh  40  and  60  lbs.  respectively.  The  co- 
efficient of  friction  under  A  is  0.2;  that  under  5  is  0.25.  When  P  =  lbs.,  the 
forces  on  the  spring  connecting  the  bodies  have  what  value? 


^VWV 


ww\ 


'vtttttttttttttttttttttttttttttpttttmtp 
Fig.  158 


Fig.  159 


3i6-(3i).     The  box  shown  in  Fig.  158  weighs  120  lbs.  and  the  body  A  80  lbs.;  the 
system  is  moving  to  the  right  on  a  floor  from  an  initial  impulse.     The  coefficient  of 
friction  between  A  and  the  box  is  zero  and  that  between  the  box  and  floor  is 
Find  the  forces  on  the  (like)  springs. 

3i7-(3i).     A  body  weighing  24  lbs.  is  projected  up  a  30  deg.  incline  at  a  velocity  of 
20  ft/sec.     The  coefficient  of  friction  between  body  and  incUne  is  .     Find  the 

position  of  the  body  after  5  sec;  after  20  sec 

3i8-(3i).     A  (Fig.  159)  weighs  160  lbs.  and  B  lbs.     The  coefficient  of  kinetic 

friction  between  A  and  the  incUne  is         ,  and  between  B  and  the  incline  0.25.    De- 


368 


(Consider  the  rope  very  flexible  and  neglect 


termine  the  acceleration  of  the  system, 
tho  mass  of  rope  and  pulley,) 
3i9-(3i)-    A,  B,  and  C  (Fig.  i6o)  weigh  60,  30,  and  10  lbs.  respectively,  and  the 
coefl&cient  of  friction  between  A  and  D  is         .    Neglect  the  stiff- 
ness of  the  rope,  its  mass,  and  that  of  the  sheave.     Find  the  ten- 
sion in  the  rope  between  B  and  C. 

32o-(3i).  The  apparatus  in  Fig.  218  (page  119)  is  being  used  to 
compress  air.  The  piston  is  10  ins.  in  diameter  and  weighs  40  lbs.; 
the  lengths  of  crank  and  connecting  rod  are  3  and  10  ins. 
respectively;  the  crank  rotates  at  100  r.p.m.;  the  pressure  on  the  top  of  piston  is 
50  lbs/in-.     Find  the  stress  in  the  top  of  the  piston  rod  when  6  =  deg. 

3  2 1-(3  2) .    A  particle  is  moving  at  a  constant  speed  of  2  ins/sec  along  an  ellipse  whose 
axes  are  and  ins.  long.     Determine  the  amount  and  direction  of  the  acceler- 

ation of  the  particle  at  the  instant  it  is  passing  one  end  of  the  er  axis. 

322-(33).     P  (Fig.  266)  moves  in  the  circle  (diameter  =  ft.)  with  constant 

speed  of  one  revolution  in  sec.      When  the  angle  POX  is  deg.,  what  are 

Vx,  i'„,  Cr,  and  «„? 


FlG.  160 


"x,   ^-y,  "J, 


U- 


323-(33).  A  projectile  is  discharged  from  a  gun  with  a  velocity  of  644  ft/sec,  at 
an  elevation  of  30  deg.  Neglecting  the  resistance  of  the  air,  compute  (o)  the  extreme 
range  of  the  projectile  measured  on  a  horizontal  plane  through  the  muzzle  of  the  gun, 
and  Q})  the  maximum  elevation  of  the  projectile  above  this  plane.  (U.  S.  Civil  Service 
examination.) 

324-(34).     A  smooth  sphere  weighing  2  lbs.  (Fig.  161)  is  hi  a  box  rigidly  fastened 
to  an  arm  which  can  be  rotated  about  a  horizontal  axis.     By  means 
of   suitable  forces  applied  to  the  arm  the  system  is  made  to  get  up 
speed  uniformly  so  that  in  every  second  the  speed  is  increased  by 
2  rev /sec.     At  the  instant  when  the  speed  reaches  rev /sec,  the  arm  is 

inclined  as  shown.  What  are  the  values  of  all  the  forces  acting  on 
the  sphere  then?  Represent  them  in  a  separate  sketch  of  the 
sphere. 

32  5-(34).     CD  (Fig.  162)  is  a  rough  incUned  plane  rotating  about  the 
vertical  axis  AB  at  a  constant  speed  of  200  rev/min.     The  body  E 
weighs  icx)  lbs.  and  its  center  of  gravity  is  3  ft.  from  AB.     Determine  the  friction 
and  normal  pressure  on  E. 

326-(34).  AB  (Fig.  163)  is  a  board  lying  upon  a  table.  C  is  a  vertical  peg  in  the 
table  top  projecting  upward  through  a  suitable  hole  in  the  board.  The  board  weighs 
20  lbs.  The  table  top  (and  board)  are  spun  about  C  at  400  rev/min.  Determine  the 
stress  at  the  smallest  section  of  the  board. 


/or-  -1 

A 

zk     ^ 

<:-  — 

\  ---.>U-   A 

Fig.  162  Fig.  163 

A  table  with  a  circular  top  has  a  smooth  border  along  the  edge  and 


327-(34) 
is  mounted  so  that  it  can  be  rotated  about  the  vertical  through  its  center. 


Suppose 


p 


369 

that  a  straight  rod  lies  on  the  top.  When  the  table  is  rotated  the  rod  rolls  to  and  its 
ends  bear  against  the  border.  Take  the  diameter  of  the  top  —  9  ft.,  length  of  rod  = 
6  ft.,  weight  of  rod  =  10  lbs.,  and  speed  of  rotation  =  200  rev/min.  Compute  the 
pressures  at  the  ends  of  the  rod. 

328-(34).     Fig.  164  represents  a  block  A  upon  which  is  mounted  a  drum  B.     A 
light  rope  is  wound  around  the  drum  so  that  a  horizontal 
pull  P  can  be  applied  as  shown.     The  weight  of  A  is  960, 
and  that  of  B  is  640  lbs.     The  coefficient  of  friction  under 
A  is  ..    When  P  =         lbs.,  the  acceleration  of  A  has  ''/h;m;w//wwh///w- 

what  value?  I'^i^-  164 

329-(34).  A  Ipcomotive  is  running  along  a  level  track.  The  horizontal  component 
of  the  reactions  of  the  rails  on  the  drivers  are  forward  or  back  according  to  certain 
circumstances;  on  the  other  wheels  it  is  opposite  to  the  direction  of  motion.  Call  the 
first  force  P;  the  second  Q.  What  is  the  relation  between  P  and  Q  (i)  when  the  loco- 
motive speed  is  constant?  (//)  when  it  is  being  increased?  (iii)  when  it  is  being 
lessened?  What  does  a  locomotive  do  to  a  bridge  over  which  it  is  moving  in  the  three 
cases? 

33o-(34)-     What  is  the  nature  of  the  action  of  a  travelling  crane  (Fig.  165)  on  its 

i:^^;^A track  (a)  when  it  is  starting  to  "travel"  (run 

down  the  yard)?  (b)  When  it  is  starting  to 
"traverse"  (crab  A  runs  over  the  bridge  B)? 
(c)  When  hoisting  of  the  load  begins? 

33i-(34).     Take  weights  of  parts  of  crane 
(preceding   problem)    as   follows:   bridge   A, 

^^7777T7777777777m777777777777m7777777/77!7^^77777777h-  tOUS;     Crab     B,  tOnS ;     load      C, 

^^°*  ^^5  tons.     Take  acceleration  of  travel  0.7  ft/sec/ 

sec,  and  acceleration  of  traverse  0.4  ft/sec/sec;  then  calculate  the  wheel  reactions 
on  track  so  far  as  possible  for  commencements  of  travel  and  traverse.  (Neglect 
swing  of  load.    What  is  effect  of  this  error?) 

332-(34).  The  following  is  an  extract  from  a  description  of  the  Sheepshead  Bay 
Motor  Racetrack  (Engineering  News  for  Aug.  19,  1915):  "There  are  two  parallel 
stra?ghtaway  stretches  connected  by  two  turns  of  180  deg.  each.  Each  turn  con- 
sists of  a  circular  arc  of  about  135  deg.  connected  by  'spirals'  to  the  straightaway 
stretches;  the  radius  to  the  inner  edge  of  the  circular  track  is  850  ft.  The  outer  edges 
of  the  circular  turns  are  given  a  maximum  super  elevation  of  25  ft.  6  in.,  computed 
for  a  speed  of  96  mi/hr  by  the  common  railway  formula  C  =  dv^/gR.  The  width  d 
was  taken  in  14  ft.  strips  commencing  at  the  inside  of  the  track,  and  the  super  ele- 
vation computed  for  speeds  of  40,  52I,  65,  yyf ,  and  96  mi/hr.  This  gives  a  cross 
section  theoretically  of  a  parabolic  curve."  Prove  the  last  statement,  and  show  how 
the  formula  gives  25  ft.  6  in. 

333-(35).  Imagine  the  surface  C  (Fig.  56)  tilted  so  that  the  top  of  B  is  horizontal; 
assume  the  angle  of  tilt  to  be  deg.,  the  surface  C  to  be  frictionless;  that  A  and  B 
respectively  weigh  and  lbs.  When  the  body  B  is  released  (P  =  o),  ^  and  B 
slide  down  the  plane  together.  Then  the  pressure  between  A  and  B  has  what  value 
and  direction? 

334~(3S)-  Fig-  166  represents  a  bar  which  rests  upon  a  horizontal  surface  in  the 
plane  of  the  paper.  The  bar  weighs  160  lbs.  and  the  coefficient  of  friction  between  the 
bar  and  table  is  0.5.     The  two  forces  shown  acting  on  the  bar  are  horizontal.     De- 


370 


^P=  100  lbs. 


=500  Ibs^ 


I  

Fig.  i66 


B 


termine  the  magnitude,  direction  and  point  of  application  of  a  third  horizontal  force, 
which,  when  acting  with  the  two  forces  shown,  will  give  the  bar  a  motion  of  trans- 
lation with  an  acceleration  to  the  right  along  AB  of 
6o  ft/sec/sec. 

33S~(35)-      B  (Fig.    167)  is   a  straight  post  which  rests 
upon  the  front   edge  of  the   car    A. 
With   what    acceleration  must  .1    be 
made  to  move  along  a  level  track  in 
order  that  B  will  remain  in  the  position  shown? 

336-(35).  Assume  that  the  car  of  the  preceding  problem 
moves  up  a  30  deg.  incline  with  a  uniform  acceleration  of  ft/sec/sec.  At  what 
angle  with  the  vertical  must  the  post  be  inclined  in  order  that,  as  before,  it  may 
maintain  its  position? 

337-(35)-  A  homogeneous  cylinder  weighing  100  lbs.  is  drawn  up  an  inclined  plane 
by  a  force  of  200  lbs.  acting  parallel  to  the  plane.  The  cylinder  is  4  ft.  long  and  2  ft. 
in  diameter;  it  rests  on  end  with  its  length  normal  to  the  plane.  The  coefficient  of 
friction  between  cyHnder  and  plane  is  0.2;  the  angle  of  inclination  of  the  plane  is 
deg.  to  the  horizontal.  Determine  the  limits  between  which  the  point  of  application 
of  the  200  lb.  force  must  lie  in  order  that  the  cylinder  may  not  tip  over. 

338-(36).  A  solid  piece  of  cast  iron  consists  of  a  right  circular  cylinder  4  ft.  in 
diameter  and  10  ft.  long,  and  a  right  circular  cone  4  ft.  in  diameter  and  ft.  long, 
placed  end  to  end.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  the 
body  with  respect  to  the  common  axis  of  cone  and  cylinder. 

339^(36).  The  moment  of  inertia  of  a  sphere  with  respect  to  a  diameter  is  given  by 
i  Mr-.  What  is  the  moment  of  inertia  of  a  cast-iron  sphere,  ins.  in  diameter,  with 
respect  to  a  tangent  line? 

340-(36).     Determine  the  moment  of  inertia  and  radius  of  gyration  of  the  cast-iron 

pulley  represented  in  Fig.  168  with  respect  to 
its  own  axis. 

34i-(37).  "The  16-inch  gun  has  a  range 
of  about  42.4  miles,  and  it  travels  this  dis- 
tance in  about^^5  seconds.  If  the  gun  were 
pointed  exactly  ndrth  or  south  the  lateral  de- 
viation of  the  projectile,  due  to  the  earth's 
rotation,  would  amount  to  525  feet."  (5a- 
cntific  American  for  May  22,  191 5.)  Show 
how  to  calculate  the  stated  deviation. 

342-(37).     The  disk  of  a  certain  steam  tur- 
FiG.  168  bine  is  ft.  in  diameter,  and  it  is  run  at 

rev/min.  What  is  the  rim  speed  in 
miles  per  hour?  Compare  the  acceleration  of  a  point  on  the  rim  with  gravity.  A 
small  bolt  is  screwed  radially  into  the  disk  at  the  rim;  compare  the  tension  in  the 
shank  of  the  bolt  just  under  the  head  with  the  weight  of  the  head . 

343-(37).  -4  and  B  (Fig.  159)  weigh  160,  and  lbs.  respectively.  The  weight 
of  the  pulley  is  lbs.,  its  diameter  is  6  ft.,  and  its  radius  of  gyration  2  ft.     The 

coefficient  of  friction  under  .-1  is  ;  under  B  it  is  \.     Determine  the  acceleration  of 

A  and  B  and  the  tension  in  both  parts  of  the  rope. 

344-(37).     Fig.  169  represents  a  drum  with  a  brake  attachment.     A  rope  is  wound 


371 


around  the  drum  and  a  load  of  2000  lbs.  hung  as  shown.     The  force  P  on  the  brake 
arm  is  1000  lbs.     The  coefficient  of  friction  between  brake  and  drum  is  ;  the 

diameter  of  the  drum  is  6  ft.,  its  weight  is  1800  lbs.,  and  its      _  ,^, >k>^^'>|| 

radius  of  gyration  is   2.5  ft.     The  load  is   released   and    i__  i     ^=J 


1 


l-rr/^) 


mc. 


w 

169 


allowed  to  fall  until  the  drum  has  attained  a  speed  of  100 
rev/min.  Then  the  brake  is  applied  and  the  system  is 
brought  to  rest.  How  much  time  is  required  for  the 
braking? 

345-(37)-  The  drum  shown  in  Fig.  170  rests  upon  a 
floor  and  against  a  low  vertical  wall  as 
shown.     A  rope  is  wound  around  the 

axle  and  passes  off  horizontally  over  the  wall.  The  diameter 
of  the  drum  is  4  ft.;  the  diameter  of  the  axle  is  3  ft.;  the  radius 
of  gyration  of  the  drum  and  axle  is  20  ins.,  and  their  combined 
weight  is  180  lbs.  The  coefficient  of  friction  between  drum 
and  floor  is  and  between  drum  and  wall  0.2.     The  pull  P  =  200  lbs.     Deter- 

mine the  angular  acceleration  of  the  drum,  and  all  forces  which  act  upon  it. 

346-(37).  Fig.  171  represents  a  brake  device  for  regulating  the  speed  at  which  a 
load  is  lowered.  It  consists  of  a  ring  solidly  fixed  in  a  horizontal  plane,  through  the 
center  of  which  passes  a  vertical  shaft.  This  shaft  carries  a  cross- 
arm  upon  which  slide  two  blocks  -4.-1.  The  rope  which  carries  the 
load  passes  over  a  pulley  and  is  wound  around  the  shaft  as  shown. 
When  the  load  starts  to  descend,  the  shaft  rotates  and  the  blocks  AA, 
shding  out  to  the  ends  of  the  cross-arm,  bear  against  the  inner  side 
of  the  ring.  The  weight  of  each  of  the  blocks  is  64  lbs.,  the  distance 
from  the  axis  of  the  shaft  to  the  center  of  gravity  of  each  block  when 
bearing  against  the  ring  is  20  ins.,  the  diameter  of  the  shaft  is  6  ins.,  the 
coefficient  of  friction  between  block  and  ring  is  ,  and  the  load  W 

is  1600  lbs.     Neglecting  the  weight  of  the  vertical  shaft   and   the 
pulley,  find  the  maximum  velocity  which  the  load  will  attain  in  descending. 

347-(38)-  Referring  to  Prob.  343:  enumerate  the  forces  acting  on  the  pulley. 
Tell  how  the  reaction  of  the  pulley  shaft  is  related  to  the  other  forces  acting  on  the 
pulley. 

348-(38).  Referring  to  Prob.  344:  determine  the  value  of  the  axle  reaction  on  the 
drum  during  the  braking  period. 

349-(38).  The  frame  (Fig.  162)  can  be  rotated  about  the  vertical  shaft  AB.  The 
shaft  is  12  ft.  long;  AD  ^  %  ft.,  CD  =  10  ft.,  and  BC  =  2  ft.  The  weights  of  these 
members  are  respectively  500,  200,  and  400  lbs.  72  is  2  X  4  ft.,  and  perpendicular  to 
paper,  i  ft.;  it  weighs  300  lbs.,  and  is  placed  at  mid-length  of  CD.  The  entire  system 
is  rotated  at  1800  rev/min.     Determine  all  forces  on  each  member. 

35o-(3o)-  The  length  of  a  simple  seconds  pendulum  at  a  certain  place  is  3.56  ft. 
Find  the  length  of  a  pendulum  which  at  the  same  place  swings  from  one  side  to  the 
other  in  5  sees.     (U.  S.  Civil  Service  examination.) 

3Si-(4o).  The  body  C  (Fig.  76)  is  moved  up  the  plane  by  a  horizontal  force  P 
(=  lbs.)  and  Q  (=         lbs.).     The  frictional  resistance  is  10  lbs.;  the  incUnation 

of  the  plane  is  deg.,  and  the  body  weighs  40  lbs.  Compute  the  work  done  on  C 
by  each  force  during  a  displacement  from  A  to  B,  20  ft.  What  is  the  algebraic  sum 
of  these  works  ("total  work")? 


372 


352-(4o).     C  (Fig.  172)  is  a  bead  on  a  circular  wire  ABD;  it  is  subjected  to  four 
forces  F,  P,  Q,  and  5;  F  =  10  lbs.  and  is  always  horizontal;  P  =  40  lbs.  and  is  always 

directed  toward  D;  Q  varies  in  magnitude  and  direction 
so  that  the  simultaneous  values  of  a,  /3,  and  Q  are  as 
follows: 

60° 


15 


30 


45 


75 


90 


/3  = 

Q  = 


5  is  a  tangential  pull  and  its  value  (in  pounds)  is  40  s~, 
where  5  is  the  arc  AC  in  feet.  Compute  the  amount  of  work 
done  by  each  force  for  the  displacement  of  C  from  A  to  B. 

3S3-(4o).  ABC  is  a.  right  triangle;  AC  is  the  hypothenuse,  AB  =  8  and  BC  =  6 
ft.  A  certain  small  body  is  made  to  move  along  AB  by  several  forces,  one  of  which  is 
always  directed  toward  C  and  equals  10  lbs.  How  much  work  does  this  force  do  while 
the  body  is  made  to  move  from  A  to  B? 

354-(4i).  When  the  pulley  described  in  Prob.  340  is  rotating  at  200  rev/min.,  what 
is  its  kinetic  energy  (in  foot-pounds)? 

35S-(4i).  When  a  solid  (circular)  cylinder  is  rolling  on  a  straight  roadway,  what 
portion  of  its  total  kinetic  energy  is  "translational"? 

356-(42).  Two  tests  were  run  on  a  certain  steam  engine.  In  the  first  test  the 
fly-wheel  spokes  were  exposed  to  the  air;  in  the  second  test  they  were  enclosed  so  as  to 
reduce  the  air  resistance  as  much  as  possible.  The  first  test  gave  an  indicated  power 
of  12.30  Continental  horse-power;  the  second  of  7.88  Continental  horse-power. 
Assuming  that  energy  is  worth  one  cent  per  kilowatt  hour,  find  the  gross  (money) 
saving  due  to  enclosing  the  fly-wheel. 

357-(42).  What  power  is  required  to  move  a  block  weighing  1200  lbs.  up  a  30  deg. 
incline  at  a  uniform  speed  of  100  ft /sec  if  the  coefficient  of  friction  between  block  and 
plane  is  0.2? 

358-(42).  S  and  S'  (Fig.  173)  are  two  portions  of  a  shaft.  Arms  A  and  A'  are 
rigidly  attached  to  the  adjacent  ends  of  the 
shaft  as  shown.  The  ends  of  the  arms  are 
furnished  with  hooks  which  are  connected  by 
two  like  coil  springs  as  shown.  Thus  it  is  pos- 
sible to  transmit  energy  from  one  portion  of  the 
shaft  to  the  other;  indeed  the  device  illus- 
trates, in  principle,  a  "transmission  dynamom- 
eter." Let  length  of  each  arm  =  ins., 
natural  length  of  each  spring  =  8  ins.,  stiffness 
of  each  spring  =  40  lbs/in.  (40  lbs.  pull  required 
for  each  inch  of  stretch).  When  the  shaft  is 
rotating  at  200  rev/min,  the  angle  between  the  arms  is 
power  of  transmission? 

359-(42).  Put  your  solution  of  the  preceding  problem  into  general  terms,  using 
the  following  notation:  a  =  length  of  each  arm  in  inches,  b  =  natural  length  of 
each  spring  in  inches,  p  =  stiffness  of  spring  in  pounds  per  inch,  n  =  speed  in  revolu- 
tions per  minute,  d  =  angle  between  arms  in  degrees. 

36o-(43).  A  flywheel  of  a  4  h-p  riveting  machine  fluctuates  between  60  and  90 
r.p.m.    Every  two  seconds  an  operation  occurs  which  requires  |  of  all  the  energy  sup- 


deg. 


Fig.  173 

What  is  the  horse- 


373 

plied  for  two  seconds.  Find  the  moment  of  inertia  of  the  wheel.  (U.  S.  Civil  Service 
examination.) 

36i-(43).  A  particle,  under  the  action  of  gravity  alone,  moves  from  rest  from  the 
highest  point  on  the  outer  surface  of  a  smooth  sphere  whose  diameter  is  lo  ft. 
Neglecting  friction  and  the  small  force  necessary  to  start  the  particle,  find:  (a)  at 
what  point  the  particle  leaves  the  sphere,  {b)  what  the  velocity  is  at  the  instant  of 
leaving.     (U.  S.  Civil  Service  examination.) 

362-(43).  A  lo-inch  rifle  has  a  barrel  45  ft.  long  and  shoots  a  shell  weighing  800 
lbs.;  the  cross  section  area  of  the  bore  is  80  in-.  The  powder  pressure  varies  from 
50,000  lbs/in-  at  the  instant  of  detonation  to  5000  lbs/in-  when  the  projectile  leaves 
the  muzzle.  Assume  the  variation  of  pressure  to  be  uniform  with  respect  to  the  dis- 
placement of  the  projectile,  and  neglect  the  effect  of  friction  and  recoil;  then  deter- 
mine the  maximum  power  developed  during  a  discharge,  and  the  muzzle  velocity  of 
the  projectile. 

363-(43).  A  car  coasts  down  a  2  per  cent  grade,  starting  from  a  point  1000  ft. 
from  the  bottom.  As  soon  as  the  level  track  is  reached  the  brakes  are  set,  locking  the 
wheels.  The  total  weight  of  the  car  is  3200  lbs.  There  are  two  pairs  of  wheels,  each 
pair,  with  the  axle,  weighing  320  lbs.,  and  having  a  radius  of  gyration  of  i  ft.;  the 
diameter  of  the  wheels  is  3  ft.  The  total  rolling  resistance  is  10  lbs.  and  the  coeffi- 
cient of  friction  between  wheels  and  track  is  0.2.  Determine  how  far  along  the  level 
track  the  car  will  go  after  the  brakes  are  set. 

364-(43).  A  flying  airplane  is  subjected  to  two  external  forces  —  gravity  and  air 
pressure.  It  is  convenient  for  purposes  of  analysis  to  regard  the  latter  in  the  three 
parts  which  act  on  the  wings,  the  propeller,  and  the  remainder  of  the  machine.  In 
"simple  flight  "  (horizontal  or  inclined  straight  path,  in  still  air  or  directly  with  or 
against  the  wind),  the  air  reaction  on  the  wings  is  dealt  with  in  two  components  — 
perpendicular  and  parallel  to  the  line  of  flight;  the  first  is  called  "lift"  L,  and  the 
second  "drag"  (formerly  "drift")  D.  The  reaction  of  the  air  on  the  propeller  can  be 
regarded  as  a  single  forward  force  or  "thrust"  T,  and  a  couple  C  opposing  the  rotation 
of  the  propeller.  The  third  force  or  "body  resistance"  R  is  directed  nearly  along  the 
Une  of  flight.     (See  Fig.  174.) 


Fig.  174 

The  lift  and  drag  depend  upon  velocity  v  of  the  airplane  relative  to  the  air,  wing 
area  A,  density  of  air  8,  "angle  of  incidence"  a  (angle  between  chord  of  wing  and 
direction  of  wind,  type  of  wing  section,  and  "aspect  ratio"  (ratio  of  length  to  breadth 
of  wing);  and  in  biplanes,  on  the  "gap"  (distance  between  the  planes)  and  to  a  lesser 
extent  on  other  details.    Thus 

L  =  KdAv'^        and        D  =  kbAv^, 


374 

where  K  and  k  are  experimental  (lift  and  drag)  coefficients.  They  are  "absolute" 
or  abstract  coefficients,  permitting  the  use  of  any  systematic  units  in  the  formulas; 
for  example,  pound  for  force,  square  foot  for  area,  foot  per  second  for  velocity,  slug 
per  cubic  foot  for  density.  (The  formulas  are  also  written  without  the  density  factor, 
the  coefficients  being  made  to  suit  certain  convenient  units  like  the  pound,  square 
foot,  and  mile  per  hour.)  The  body  resistance  varies  approximately  as  the  velocity 
square,  size  and  shape  of  body,  and  other  structural  details.  In  any  case,  the  formula 
can  be  written  R  =  Cv^,  where  C  is  a  coefficient  which  can  be  estimated  by  the  expert 
designer. 

In  horizontal  flight  the  weight  of  the  airplane  is  balanced  by  the  lift  and  the  vertical 
component  of  the  thrust.  But  this  component  is  neglected  in  ordinary  calculations. 
Thus  we  have 

KhAv"  =  W        or        v'^=W^  K5A. 
In  uniform  horizontal  flight  the  total  resistance  D  +  R  and  the  thrust  T  are  equal, 
and  the  work  done  against  the  total  resistance  per  unit  time  and  the  thrust  power  are 
equal.     If  the  airplane  is  moving  in  still  air,  this  thrust  or  useful  power  is 

P  =  Tv  =  {D  +  R)v. 

The  following  data  refer  to  a  particular  monoplane:  weight  =  looo  lbs.;  area  of 
wings  =  155  ft^;  C  =  0.02;  and  for  angle  of  incidence 

a  =  —2  o  2  4  6  8  10         12         14         16      18  deg. 

K  =  0.000  0.070  0.150     0.235  0.305  0.380  0.435  0-475  0-495  o-5oo     0.480 

k  =  0.017  0.014  0.014     0.017  0.024  0.033  0.043  0.052  0.063  0.078     O.IOO 

(i)  Plot  values  of  K,  k,  and  K/k  on  an  «  base,  using  scales  of  i  in.  =  4  deg.,  and  i 
in.  =  o.i. 

(ii)   Compute  values  of  v  for  the  stated  values  of  a. 

(iii)  Compute  values  of  D  and  R  for  the  computed  values  of  v.  Plot  values  of  D,  R, 
and  D  -\-  R  on  the  same  v  base,  using  scales  i  in.  =  20  ft/sec,  and  i  in.  =  100  lbs. 
(iv)  Compute  values  of  P  for  the  computed  values  of  v.  Reduce  these  values  of  P 
to  horse-power,  and  then  plot  horse-power  required  on  the  velocity  base  used  in  (iii). 
(The  horse-power  available  plotted  on  a  velocity  base  gives  a  curve  convex 
upward,  intersecting  the  horse-power  required  curve  in  two  points.  The  velocities 
corresponding  to  these  two  points  are  the  minimum  and  maximum  velocities  of  hori- 
zontal ffight  for  the  airplane.) 

365-(43).  When  the  motor  of  a  flying  airplane  is  cut  out,  the  plane  soon  takes  on  a 
gUde  at  an  angle  and  speed  depending  mainly  on  the  set  of  the  elevator.  Then  the 
propeller  is  driven  by  the  thrust  which  is  now  opposed  to  the  motion  of  the  airplane; 
but  this  thrust  is  small  compared  to  the  drag.     Show  that 

where  d  =  the  angle  of  the  glide.  Compute  values  of  9  and  v  for  the  values  of  a  given 
in  the  preceding  problem,  and  then  plot  cot  6  on  the  v  base. 

366-(43).  A  locomotive  weighing  200  tons  pulls  a  train  of  30  cars,  each  weighing 
50  tons.  At  10  mi/hr  the  locomotive  resistance  is  10  lbs/ton,  and  the  train  resistance 
is  4  lbs/ton.  What  net  horse-power  is  necessary  to  maintain  that  speed  on  a  i 
per  cent  grade? 

367-(43).  Find  the  total  work  done  at  the  drawbar  of  a  locomotive  in  starting  a 
200-ton  train  against  a  one  per  cent  grade  from  rest  to  30  mi/hr  in  300  ft.  when  the 


375 


frictional  resistances  are  lo  lbs/ton  (assumed  the  same  for  all  speeds).     (Mass.  Civil 
Service  examination.) 

368-(43).  The  engine  of  a  certain  locomotive  weighs  ii8  tons,  io6  on  the  drivers; 
the  loaded  tender  weighs  60  tons.  The  cylinders  are  24  (diameter)  by  30  ins.  (length), 
and  the  drivers  are  60  ins.  in  diameter;  the  boiler  pressure  is  200  lbs/in-.  What  is  the 
highest  velocity  at  which  this  locomotive  can  haul  a  train  weighing  looo  tons  on  a 
level  track?  (Assume  that  the  train  resistance  follows  the  Engineering  News  formula, 
and  that  the  mean  effective  follows  the  law  p  =  200(0.95  —  (7  5  -i-  11,000)];  see 

Prob.  149.) 
369-(44).     Fig.  175  shows  plan  of  a  capstan  and  hauling  tackle,  and  an  elevation 


-^-A 


IS'- 


c-S" 


Arm 

i 


Fig.  17s 

of  the  barrel  of  the  capstan.  The  sweep  (arm)  to  which  the  horse  is  hitched  at  ^  is  11 
ft.  long;  there  is  one  sheave  in  each  block  B  and  C.  Assume  that  the  horse  can  exert 
a  prolonged,  steady  pull  of  lbs.,  and  make  an  estimate  (supported  by  calculation) 
of  the  pull  which  can  be  exerted  at  the  load.     (Neglect  friction  on  the  barrel.) 

37o-(44).  Fig.  176  represents  in  plan  and  elevation  a  coal  shipping  station  for 
loading  barges  at  a  Mississippi  River  "  landing."  The  coal  is  brought  in  railway  cars 
up  on  to  the  trestle  to  the  supply  track,  and  then  dumped  into  a  hopper  from  which 
it  is  discharged  as  required  into  a  transfer  car.  This  car  runs  on  a  track,  extending 
toward  the  river  and  up  the  cradle  to  a  dumping  platform,  whence  it  discharges  its 
load  into  the  barge  below. 

The  loading  plant  is  operated  by  means  of  three  cables,  two  to  handle  the  transfer 
car,  and  one  to  handle  the  cradle.  The  winding  drums  for  the  cables  are  mounted  in 
the  "operating  house,"  and  are  driven  by  gearing  from  an  electric  35  horse-power 
motor.  The  hauling  cable  A  extends  from  the  drum  A  vertically  downwards  to  a 
sheave,  thence  horizontally  to  another  sheave  in  the  middle  of  the  track,  thence  along 
the  center  of  the  track  down  the  incline  and  through  the  cradle,  passing  under  a  10- 
inch  pulley  (not  clearly  indicated)  and  a  30-inch  pulley,  and  then  up  to  a  36-inch 
pulley  at  the  track  level;  the  end  of  the  cable  is  attached  to  the  transfer  car.  The 
back-haul  cable  B  extends  vertically  down  from  drum  5  to  a  sheave,  thence  horizon- 
tally, and  by  turns  over  two  sheaves,  to  a  point  in  the  upper  end  of  the  inclined  track; 
thence  down  the  center  of  the  track  and  over  the  deck  of  the  cradle  to  the  car.  The 
cradle-cable  C  leads  from  its  drum  downwards  to  a  sheave,  thence  along  one  side  of 
the  incline  to  the  upper  end  of  the  cradle;  it  crosses  under  by  means  of  two  sheaves 
and  then  extends  along  the  opposite  side  of  the  incline  to  a  point  of  attachment  on  the 
trestle  which  carries  the  overhead  track.  Normally,  the  cradle  will  be  held  in  position 
by  a  ratchet  brake  on  the  first  pair  wheels.  Diameter  of  all  sheaves  not  given  above 
or  indicated  in  the  figure  is  30  ins.,  diameter  of  the  cables  is  H  in.  The  cradle  weighs 
about  22  tons,  the  transfer  car  empty  about  8  tons,  and  loaded  about  20  tons. 


376 


o 

t--. 


o 


377 


L 


-^ 


B 


±3 


\ — ^ 


C 


B 


3 


Fig. 


177 


Compute:  (a)  the  torque  required  at  the  drum  A  for  pulling  the  loaded  transfer 
car  up  the  cradle  from  a  "standing  start";  (6)  the  torque  required  at  the  drum  B  to 
pull  the  empty  car  up  the  5  per  cent  grade  from  a  standing  start;  (c)  the  torque  re- 
quired at  the  drum  C  for  pulling  the  cradle  up  its  track,  transfer  car  not  on  the  cradle. 
37i-(45).  A  (Fig.  177)  is  a  long  board  supported  in  a  horizontal  position;  5  is  a 
Week  provided  with  end  guides  so  that  it  can  be  sHd  along  the  board  without  turning; 

lid  C  is  a  heavy  piece  that  can  be  slid  across  the  board. 
Suppose  that  B  —  and  with  it  C  —  is  moved  to  the  left  with 
a  velocity  of  4  ft /sec,  and  C  is  moved  across  the  board,  in 
the  direction  indicated,  with  a  velocity  of  0.2  ft /sec.  (0) 
What  is  the  direction  of  the  friction  exerted  by  .1  on  C? 
{b)  What  is  the  amount  of  that  force  if  the  weight  of  C  — 
all  supported  hy  A  —  is  10  lbs.,  and  the  coefficient  of  ki- 
netic friction  between  A  and  B  is  0.4?  (c)  If  the  coefficient 
between  B  and  C  is  0.4,  what  is  the  amount  of  friction 
between  them?  (</)  Compare  the  total  friction  on  C  just  found,  with  its  value 
when  C  is  slid  across  but  not  also  along  the  board?  (c)  Make  a  similar  comparison 
on  the  supposition  that  the  contact  between  B  and  C  is  frictionless. 

372-(45).  Put  the  solution  of  the  foregoing  problem  into  general  terms,  denoting 
the  velocities  respectively  by  v  and  F,  the  coefficient  of  friction  by  ix,  the  weight  of  C 
hyW. 

373-(45)-  "To  easily  push  a  pulley  onto  a  snugly  fitting  arbor  or  shaft,  one  turns 
the  pulley  while  pushing."     Explain  how  the  turning  lessens  the  necessary  push? 

374-(45).  To  turn  a  certain  pulley  on  a  snugly  fitting  shaft,  3  inches  in  diameter, 
requires  a  torque  of  40  ft. -lbs.  How  large  a  pull  is  required  to  strip  the  pulley  from 
its  shaft  at  a  speed  v  when  the  wheel  is  being  turned  so  that  the  rate  of  circumferential 
slip  is  F? 

375-(45)-     Fig-  178  represents  a  cone  friction  clutch,  used  in  some  automobiles. 

A  is  the  engine  flywheel;  the  inner  face  of  the 
rim  is  conical.  B  is  the  other  cone  keyed  to  the 
transmission  shaft  but  so  that  B  can  be  slid  along 
the  shaft,  toward  or  from  the  cone  A.  Generally, 
the  clutch  is  "thrown  in"  {B  pressed  against  .4) 
when  the  engine  (and  flywheel)  are  running. 
The  cones  slip  over  each  other,  the  (kinetic)  fric- 
tion on  B  increasing  the  speed  of  B  until  its  speed 
reaches  that  of  A .  Thereafter  the  friction  is  static. 
Suppose  that  the  engine  can  deliver  a  torque  T 
at  the  flywheel  at  a  certain  speed  of  running.  To 
transmit  that  torque  the  clutch  must  develop  a 
circumferential  friction,  T  -^  ^  d;  and  hence  a  nor- 
mal pressure  2  T  ^  nd.  Show  that  the  force  re- 
quired for  pressing  the  cones  together  is  between 

2  r  sin  0  -^  ^rf    and    2  T  (sin  0  +  m  cos  6)  -i-  /ud. 
376-(45).     Solve  Prob.  369,  making  an  allowance  for  friction  on  the  barrel. 


378 


NATURAL  SINES 


SINES,  angles  0°  to  45°.    Example,  sin  33.3°  =  0.5490 


Angle 

°.o 

°.l 

°.2 

°.3 

°.4 

°.5 

°.6 

°.7 

°.8 

°.9 

0.0000 

90° 

0° 

o.oooo 

0017 

0035 

0052 

0070 

0087 

oios 

0122 

0140 

0157 

OI7S 

89 

17 

1 

OI75 

0192 

0209 

0227 

0244 

0262 

0279 

0297 

0314 

0332 

0349 

88 

17 

2 

03  49 

0366 

0384 

0401 

0419 

0436 

0454 

0471 

0488 

0506 

0523 

87 

17 

3 

0523 

0541 

0558 

0576 

0593 

0610 

0628 

064s 

0663 

0680 

0698 

86 

17 

4 

0698 

0715 

0732 

0750 

0767 

0785 

0802 

0819 

0837 

0854 

0.0872 

85 

17 

6 

0.0872 

0889 

0906 

0924 

0941 

0958 

0976 

0993 

lOII 

1028 

104s 

84 

17 

6 

1045 

1063 

1080 

1097 

1115 

1132 

1 149 

1167 

1184 

1201 

1219 

83 

17 

7 

1219 

1236 

1253 

1271 

1288 

1305 

1323 

1340 

1357 

1374 

1392 

82 

17 

8 

1392 

1409 

1426 

1444 

1461 

1478 

1495 

1513 

1530 

1547 

1564 

81 

17 

9 

1564 

1582 

1 599 

1616 

1633 

1650 

1668 

1685 

1702 

1719 

0.1736 

80° 

17 

10° 

0.1736 

I7S4 

1771 

1788 

1 80s 

1822 

1840 

I8S7 

1874 

1891 

1908 

79 

17 

11 

1908 

1925 

1942 

1959 

1977 

1994 

201 1 

2028 

2045 

2062 

2079 

78 

17 

12 

2079 

2096 

2113 

2130 

2147 

2164 

2181 

2198 

2215 

2233 

2250 

77 

17 

13 

2250 

2267 

2284 

2300 

2317 

2334 

2351 

2368 

2385 

2402 

2419 

76 

17 

14 

2419 

2436 

2453 

2470 

2487 

2504 

2521 

2538 

2554 

2571 

0.2588 

75 

17 

16 

0.2588 

2605 

2622 

2639 

2656 

2672 

2689 

2706 

2723 

2740 

2756 

74 

17 

16 

2756 

2773 

2790 

2807 

2823 

2S40 

2857 

2874 

2890 

2907 

2924 

73 

17 

17 

2924 

2940 

2957 

2974 

2990 

3007 

3024 

3040 

3057 

3074 

3090 

72 

17 

18 

3090 

3107 

3123 

3140 

3156 

3173 

3190 

3206 

3223 

3239 

3256 

71 

17 

19 

3256 

3272 

3289 

3305 

3322 

3338 

3355 

3371 

3387 

3404 

0.3420 

70° 

16 

20° 

0.3420 

3437 

3453 

3469 

3486 

3502 

3518 

3535 

3551 

3567 

3584 

69 

16 

21 

3584 

3600 

3616 

3633 

3649 

366s 

3681 

3697 

3714 

3730 

3746 

68 

16 

22 

3746 

3762 

3778 

3795 

3811 

3827 

3843 

3859 

3875 

3891 

3907 

67 

16 

23 

3907 

3923 

3939 

3955 

3971 

3987 

4003 

4019 

4035 

4051 

4067 

66 

16 

24 

4067 

4083 

4099 

4115 

4131 

4147 

4163 

4179 

419s 

4210 

0.4226 

65 

16 

25 

0.4226 

4242 

4258 

4274 

4289 

4305 

4321 

4337 

4352 

4368 

4384 

64 

16 

26 

4384 

4399 

4415 

4431 

4446 

4462 

4478 

4493 

4509 

4524 

4540 

63 

16 

27 

4540 

4555 

4571 

4586 

4602 

4617 

4633 

4648 

4664 

4679 

4695 

62 

16 

28 

4695 

47IO 

4726 

4741 

4756 

4772 

4787 

4802 

4818 

4833 

4848 

61 

15 

29 

4848 

4863 

4879 

4894 

4909 

4924 

4939 

4955 

4970 

4985 

0.5000 

60° 

15 

30° 

0.5000 

501S 

5030 

5045 

5060 

5075 

5090 

5105 

5120 

5135 

5150 

59 

IS 

31 

5150 

5165 

5180 

5195 

5210 

5225 

5240 

5255 

5270 

5284 

5299 

58 

IS 

32 

5299 

5314 

5329 

5344 

5358 

5373 

5388 

5402 

5417 

5432 

5446 

67 

IS 

33 

5446 

5461 

5476 

5490 

SS05 

5519 

5534 

5548 

5563 

5577 

5592 

66 

15 

34 

5592 

5606 

5621 

5635 

5650 

5664 

5678 

5693 

5707 

5721 

0.5736 

66 

14 

36 

0.5736 

S750 

5764 

5779 

5793 

5807 

5821 

5835 

5850 

5864 

5878 

54 

14 

36 

5878 

5892 

5906 

5920 

5934 

5948 

5962 

5976 

5990 

6004 

6018 

53 

14 

37 

6018 

6032 

6046 

6060 

6074 

6088 

6101 

611S 

6129 

6143 

6157 

52 

14 

38 

6157 

6170 

6184 

6198 

6211 

6225 

6239 

6252 

6266 

6280 

6293 

51 

14 

39 

6293 

6307 

6320 

6334 

6347 

6361 

6374 

6388 

6401 

6414 

0.6428 

50° 

13 

40° 

0.6428 

6441 

6455 

6468 

6481 

6 194 

6508 

6521 

6534 

6547 

6561 

49 

13 

41 

6561 

6574 

6587 

6600 

6613 

6626 

6639 

6652 

6665 

6678 

6691 

48 

13 

42 

6691 

6704 

6717 

6730 

6743 

6756 

6769 

6782 

6794 

6807 

6820 

47 

13 

43 

6820 

6833 

6845 

6858 

6S71 

6884 

6896 

6909 

6921 

6934 

6947 

46 

13 

44 

6947 

6959 

6972 

6984 

6997 

7009 

7022 

7034 

7046 

7059 

0.7071 

46° 

12 

45° 

0.7071 

°.9 

°.8 

°.7 

°.6 

°.6 

°.4 

°.3 

°.2 

M 

°.o 

Angle 

oJ 

<T3 

COSINES,  angles  45°  to  90°.    Example,  cos  66.6°  =0  3971 


AND  COSINES 


379 


SINES,  angles  45°  to  90°.    Example,  sin  66.6°  =  0.9178 


Angle 

°.o 

°.l 

°.2 

°.3 

°A 

°.5 

°.6 

°.7 

°.8 

°.9 

0.7071 

46° 

45° 

0.7071 

7083 

7096 

7108 

7120 

7133 

714s 

7157 

7169 

7181 

7193 

44 

12 

46 

7193 

7206 

7218 

7230 

7242 

7254 

7266 

7278 

7290 

7302 

7314 

43 

12 

47 

7314 

7325 

7337 

7349 

7361 

7373 

7385 

7396 

7408 

7420 

7431 

42 

12 

48 

7431 

7443 

7455 

7466 

7478 

7490 

7501 

7513 

7524 

7536 

7547 

41 

12 

49 

7547 

7559 

7570 

7581 

7593 

7604 

761S 

7627 

7638 

7649 

0.7660 

40° 

II 

60° 

0.7660 

7672 

7683 

7694 

7705 

7716 

7727 

7738 

7749 

7760 

7771 

39 

II 

61 

7771 

7782 

7793 

7804 

7815 

7826 

7837 

7848 

7859 

7869 

7880 

38 

II 

62 

7880 

7891 

7902 

7912 

7923 

7934 

7944 

7955 

7965 

7976 

7986 

37 

II 

63 

7986 

7997 

8007 

8018 

8028 

8039 

8049 

8059 

8070 

8080 

8090 

36 

10 

64 

8090 

8100 

8111 

8121 

8131 

8141 

8151 

8161 

8171 

8181 

0.8192 

35 

10 

65 

0.8192 

8202 

8211 

8221 

8231 

8241 

8251 

8261 

8271 

8281 

8290 

34 

10 

66 

8290 

8300 

8310 

8320 

8329 

8339 

8348 

8358 

8368 

8377 

8387 

33 

10 

57 

8387 

8396 

8406 

841S 

8425 

8434 

8443 

8453 

8462 

8471 

8480 

32 

9 

68 

8480 

8490 

8499 

8508 

8517 

8526 

8536 

8545 

8554 

8563 

8572 

31 

9 

59 

8572 

8581 

8590 

8599 

8607 

8616 

862s 

8634 

8643 

8652 

0.8660 

30° 

9 

60° 

0.8660 

S669 

8678 

8686 

8695 

8704 

8712 

8721 

8729' 

8738 

8746 

29 

9 

61 

8746 

8755 

8763 

8771 

8780 

8788 

8796 

880s 

8813 

8821 

8829 

28 

8 

62 

8829 

8838 

8846 

8854 

8862 

8870 

8878 

8886 

8894 

8902 

8910 

27 

8 

63 

8910 

8918 

8926 

8934 

8942 

8949 

8957 

896s 

8973 

8980 

8988 

26 

8 

64 

8988 

8996 

9003 

90H 

9018 

9026 

9033 

9041 

9048 

9056 

0.9063 

25 

7 

66 

0 . 9063 

9070 

9078 

9085 

9092 

9100 

9107 

9114 

9121 

9128 

9135 

24 

7 

66 

913s 

9143 

9150 

9157 

9164 

9171 

9178 

9184 

9191 

9198 

9205 

23 

7 

67 

9205 

9212 

9219 

922s 

9232 

9239 

9245 

9252 

9259 

9265 

9272 

22 

7 

68 

9272 

9278 

9285 

9291 

9298 

9304 

9311 

9317 

9323 

9330 

9336 

21 

6 

69 

9336 

9342 

9348 

9354 

9361 

9367 

9373 

9379 

9385 

9391 

0.9397 

20° 

6 

70° 

0.9397 

9403 

9409 

9415 

9421 

9426 

9432 

9438 

9444 

9449 

9455 

19 

6 

71 

9455 

9461 

9466 

9472 

9478 

9483 

9489 

9494 

9SOO 

9505 

9511 

18 

6 

72 

9511 

9516 

9521 

9527 

9532 

9537 

9542 

9548 

9553 

9558 

9563 

17 

5 

73 

9563 

9568 

9573 

9578 

9583 

9588 

9593 

9598 

9603 

9608 

.  9613 

16 

5 

74 

9613 

9617 

9622 

9627 

9632 

9636 

9641 

9646 

9650 

9655 

0.9659 

15 

5 

75 

0.9659 

9664 

9668 

9673 

9677 

9681 

9686 

9690 

9694 

9699 

9703 

14 

4 

76 

9703 

9707 

9711 

9715 

9720 

9724 

9728 

9732 

9736 

9740 

9744 

13 

4 

77 

9744 

9748 

9751 

9755 

9759 

9763 

9767 

9770 

9774 

9778 

9781 

12 

4 

78 

9781 

9785 

9789 

9792 

9796 

9799 

9803 

9806 

9810 

9813 

9816 

11 

3 

79 

9816 

9820 

9823 

9826 

9829 

9833 

9836 

9839 

9842 

9845 

0.9848 

10° 

3 

80° 

0.9848 

9851 

9854 

9857 

9860 

9863 

9866 

9869 

9871 

9874 

9877 

9 

3 

81 

9877 

9880 

9882 

9885 

98S8 

9890 

9893 

9895 

9898 

9900 

9903 

8 

3 

82 

9903 

9905 

9907 

9910 

9912 

9914 

9917 

9919 

9921 

9923 

9925 

7 

2 

83 

9925 

9928 

9930 

9932 

9934 

9936 

9938 

9940 

9942 

9943 

9945 

6 

2 

84 

9945 

9947 

9949 

9951 

9952 

9954 

9956 

9957 

9959 

9960 

.9962 

6 

2 

85 

0.9962 

9963 

996s 

9966 

9968 

9969 

9971 

9972 

9973 

9974 

9976 

4 

I 

86 

9976 

9977 

9978 

9979 

9980 

9981 

9982 

9983 

9984 

998s 

9986 

3 

I 

87 

9986 

9987 

9988 

9989 

9990 

9990 

9991 

9992 

9993 

9993 

9994 

2 

I 

88 

9994 

9995 

9995 

9996 

9996 

9997 

9997 

9997 

9998 

9998 

0.9998 

1 

0 

89 

0.9998 

9999 

9999 

9999 

9999 

0000 

0000 

0000 

0000 

0000 

I. 0000 

0° 

0 

90° 

I. 0000 

°.9 

°.8 

°.7 

°.6 

°.5 

°.4 

°.3 

°.2 

°.l 

°.o 

1 

Angle 

COSINES,  angles  0°  to  45°.    Example,  cos  33. S"  =  0.8358 


38o 


NATURAL  TANGENTS 


TANGENTS,  angles  0°  to  45°.    Example,  tan  33. 3°  =  0.6569 


Angle 

°.o 

M 

°.2 

°.3 

°.4 

°.5 

°.6 

°.7 

°.S 

°9 

<D 

.0000 

90° 

0" 

o . oooo 

0017 

0035 

0052 

0070 

00S7 

oios 

0122 

0140 

0157 

0175 

89 

17 

1 

OI7S 

0192 

0209 

0227 

0244 

0262 

0279 

0297 

0314 

0332 

0349 

88 

17 

2 

0349 

0367 

0384 

0402 

0419 

0437 

0454 

0472 

0489 

0507 

0524 

87 

17 

3 

0524 

0542 

0559 

OS77 

0594 

0612 

0629 

0647 

0664 

0682 

0699 

86 

18 

4 

0699 

0717 

0734 

0752 

0769 

0787 

0805 

0822 

0840 

0857 

0.0875 

85 

18 

5 

0.0875 

0892 

0910 

0928 

0945 

0963 

0981 

0998 

10 1 5 

1033 

1051 

84 

18 

6 

1051 

1069 

1086 

1 104 

1122 

1139 

1157 

1175 

1192 

1210 

1228 

83 

18 

7 

1228 

1246 

1263 

1281 

1299 

1317 

1334 

1352 

1370 

1388 

1405 

82 

18 

8 

1405 

1423 

1441 

1459 

T477 

1495 

1512 

1530 

1548 

1566 

1584 

81 

18 

9 

1584 

1602 

1620 

1638 

165s 

1673 

1691 

1709 

1727 

1745 

0.1763 

80° 

18 

10° 

0.1763 

1781 

1799 

1817 

183s 

1853 

1871 

1890 

190S 

1926 

1944 

79 

18 

11 

1944 

1962 

19S0 

1998 

2016 

2035 

20S3 

2071 

2089 

2107 

2126 

78 

18 

12 

2126 

2144 

2162 

2180 

2199 

2217 

2235 

2254 

2272 

2290 

2309 

77 

18 

13 

2309 

2327 

2345 

2364 

2382 

2401 

2419 

2438 

2456 

2475 

2493 

76 

18 

14 

2493 

2512 

2530 

2549 

2568 

25S6 

2605 

2623 

2642 

2661 

0.2679 

75 

19 

15 

0.2679 

2698 

2717 

2736 

2754 

2773 

2792 

2811 

2830 

2849 

2867 

74 

19 

16 

2867 

2886 

2905 

2924 

2943 

2962 

2981 

3000 

3019 

3038 

3057 

-,3 

19 

17 

3057 

3076 

3095 

3115 

3134 

3153 

3172 

3191 

3211 

3230 

3249 

72 

19 

18 

3249 

3269 

3288 

3307 

3327 

3346 

3365 

3385 

3404 

3424 

3443 

71 

19 

19 

3443 

3463 

3482 

3502 

3522 

3541 

3561 

3581 

3600 

3620 

0.3640 

70° 

20 

20° 

0.3640 

3659 

3679 

3699 

3719 

3739 

3759 

3779 

3799 

3819 

3839 

69 

20 

21 

3839 

3859 

3879 

3899 

3919 

3939 

3959 

3979 

4000 

4020 

4040 

68 

20 

22 

4040 

4061 

4081 

4101 

4122 

4142 

4163 

4183 

4204 

4224 

4245 

67 

21 

23 

424s 

4265 

4286 

4307 

4327 

4348 

4369 

4390 

4411 

4431 

4452 

66 

21 

24 

4452 

4473 

4494 

451S 

4536 

4557 

4578 

4599 

4621 

4642 

0 . 4663 

66 

21 

26 

0.4663 

4684 

4706 

4727 

4748 

4770 

4791 

4813 

4834 

4856 

4877 

64 

21 

26 

4877 

4S99 

4921 

4942 

4964 

49S6 

5008 

5029 

5051 

5073 

5095 

63 

22 

27 

509s 

5117 

5139 

5161 

5184 

5206 

5228 

5250 

5272 

5295 

5317 

62 

22 

28 

5317 

5340 

5362 

5384 

5407 

5430 

5452 

5475 

5498 

5520 

5543 

61 

23 

29 

5543 

5566 

5589 

5612 

5635 

5658 

5681 

5704 

5727 

5750 

0.5774 

60° 

23 

30° 

O.S774 

5797 

5820 

5S44 

5867 

5890 

5914 

5938 

5961 

598s 

6009 

69 

24 

31 

6009 

6032 

6056 

6080 

6104 

6128 

6152 

6176 

6200 

6224 

6249 

68 

24 

32 

6249 

6273 

6297 

6322 

6346 

6371 

6395 

6420 

644s 

6469 

6494 

67 

25 

33 

6494 

6519 

6544 

6569 

6594 

6619 

6644 

6669 

6694 

6720 

6745 

66 

25 

34 

6745 

6771 

6796 

6822 

6847 

6873 

6899 

6924 

6950 

6976 

0.7002 

65 

26 

35 

0 .  7002 

7028 

7054 

7080 

7107 

7133 

7159 

7186 

7212 

7239 

7265 

64 

26 

36 

7265 

7292 

7319 

7346 

7373 

7400 

7427 

7454 

7481 

7508 

7536 

63 

27 

37 

7536 

7563 

7590 

7618 

7646 

7673 

7701 

7729 

7757 

7785 

7813 

62 

28 

38 

7813 

7841 

7S69 

7898 

7926 

7954 

7983 

8012 

8040 

8069 

8098 

61 

28 

39 

8098 

8127 

8156 

8185 

8214 

8243 

8273 

8302 

8332 

8361 

0.8391 

B0° 

29 

40° 

0.8391 

8421 

8451 

8481 

8511 

8541 

8571 

8601 

8632 

8662 

8693 

49 

30 

41 

8693 

8724 

8754 

8785 

8816 

8847 

8878 

8910 

8941 

8972 

9004 

48 

31 

42 

9004 

9036 

9067 

9099 

9131 

9163 

919s 

9228 

9260 

9293 

9325 

47 

32 

43 

9325 

9358 

9391 

9424 

9457 

9490 

9523 

9556 

9590 

9623 

0.9657 

46 

33 

44 

0.9657 

9691 

9725 

9759 

9793 

9827 

9861 

9S96 

9930 

9965 

I . 0000 

45° 

34 

46° 

I. 0000 

°.9 

°.8 

°.7 

°.6 

°.5 

°.4 

°.3 

°.2 

°.l 

°.o 

Angle 

COTANGENTS,  angles  45°  to  90°.    Example,  cot  66. 6°  =  0.4327 


AND  COTANGENTS 


381 


TANGENTS,  angles  45°  to  90°.    Example,  tan  66  6°  =  2  311 


Angle 

°.0 

°.l 

°.2 

°.3 

°.4 

°.5 

°.6 

°.7 

°.8 

°.9 

i 

<T3 

[.0000 

46° 

46° 

I. 0000 

0035 

0070 

oios 

0141 

0176 

0212 

0247 

0283 

0319   0355 

41 

35 

46 

03SS 

0392 

0428 

0464 

0501 

0538 

0575 

0612 

0649 

06861  07241 

43 

37 

47 

0724 

0761 

0799 

0837 

0875 

0913 

0951 

0990 

1028 

1067   1106 

42 

38 

48 

1 106 

1 145 

1 184 

1224 

1263 

1303 

1343 

1383 

1423 

1463   1504 

41 

40 

49 

1504 

1544 

158s 

1626 

1667 

1708 

1750 

1792 

1833 

1875 

1.1918 

40° 

41 

60° 

1.1918 

i960 

2002 

2045 

2088 

2131 

2174 

2218 

2261 

2305 

2349 

39 

43 

61 

2349 

2393 

2437 

2482 

2527 

2572 

2617 

2662 

2708 

2753 

2799 

38 

45 

68 

2799 

2846 

2892 

2938 

2985 

3032 

3079 

3127 

3175 

3222 

3270 

37 

47 

63 

3270 

3319 

3367 

3416 

3465 

3514 

3564 

3613 

3663 

3713 

3764 

36 

49 

64 

3764 

3814 

3865 

3916 

3968 

4019 

4071 

4124 

4176 

4229 

I. 4281 

35 

52 

66 

I. 4281 

4335 

4388 

4442 

4496 

4550 

4605 

4659 

471S 

4770 

4826 

34 

55 

66 

4826 

4882 

4938 

4994 

5051 

5108 

5166 

5224 

5282 

5340 

5399 

33 

57 

67 

5399 

5458 

5517 

5577 

5637 

5697 

5757 

5818 

5880 

5941 

6003 

32 

60 

68 

6003 

6d66 

6128 

6191 

6255 

6319 

6383 

6447 

6512 

6577 

6643 

31 

64 

69 

1.6643 

6709 

6775 

6842 

6909 

6977 

7045 

7113 

71S2 

7251 

I. 7321 

30° 

67 

60° 

I  732 

1.739 

1.746 

1-753 

1.760 

1.767 

1-775 

1.782 

1.789 

1-797 

1.804 

29 

7 

61 

1.804 

i.8[i 

1. 819 

1.827 

1.834 

1.842 

1.849 

1.857 

1.865 

1-873 

1. 881 

28 

8 

62 

1. 881 

1.889 

1.897 

1. 90s 

I  913 

1. 921 

1.929 

1-937 

1.946 

1-954 

1.963 

27 

8 

63 

1.963 

1. 971 

1.980 

1.988 

1.997 

2  .006 

2.014 

2.023 

2.032 

2.041 

2.050 

26 

9 

64 

2.050 

2.059 

2.069 

2.078 

2.087 

2.097 

2 .  106 

2. 116 

2.125 

2.135 

2.145 

25 

9 

66 

2.14s 

2.154 

2.164 

2.174 

2.184 

2.194 

2.204 

2.215 

2.225 

2.236 

2.246 

24 

10 

66 

2.246 

2.257 

2.267 

2.278 

2.289 

2.300 

2. 311 

2.322 

2.?^3 

2.344 

2.356 

23 

II 

67 

2.356 

2.367 

2.379 

2.391 

2  .402 

2.414 

2.426 

2.438 

2.4':o 

2.463 

2.475 

22 

12 

68 

2.475 

2.488 

2.500 

2.513 

2.526 

2 .  539 

2.552 

2.56s 

2.578 

2.592 

2.605 

21 

13 

69 

2  605 

2.619 

2.633 

2.646 

2.660 

2-675 

2  689 

2.703 

2.718 

2.733 

2.747 

20° 

14 

70° 

2.747 

2.762 

2.778 

2  ■  793 

2.808 

2.824 

2.840 

2.856 

2.872 

2.888 

2.904 

19 

16 

71 

2.904 

2.921 

2.937 

2-954 

2.971 

2.989 

3.006 

3024 

3  042 

3.060 

3.078 

18 

17 

72 

3.078 

3.096 

3-II5 

3-133 

3152 

3-172 

3-191 

3. 211 

3  230 

3.251 

3.271 

17 

19 

73 

3.271 

3.291 

3-312 

3-333 

3-354 

3-376 

3-398 

3.420 

3  442 

3  465 

3.487 

16 

22 

74 

3.487 

3-511 

3-534 

3-558 

3-582 

3.606 

3-630 

3.65s 

3.681 

3  706 

3.732 

15 

24 

76 

3.732 

3.758 

3.785 

3-812 

3.839 

3  867 

3-895 

3.923 

3  952 

3.981 

4. on 

14 

28 

76 

4. on 

4.041 

4.071 

4.102 

4-134 

4-165 

4-198 

4  ■  2,30 

4.264 

4-297 

4.331 

13 

32 

77 

4.331 

4  366 

4.402 

4-437 

4-474 

4-511 

4  548 

4  586 

4.625 

4.665 

4. 70s 

12 

37 

78 

4  705 

4-745 

4-787 

4.829 

4-872 

4  91s 

4.959 

5005 

5  050 

5  097 

5-145 

11 

44 

79 

5-145 

5 -193 

5-242 

5  292 

5-343 

5-396 

5-449 

5-503 

5-558 

5-614 

5-671 

10° 

S3 

80° 

5.671 

5  -  7.^0 

5  789 

5-850 

5  912 

5 -976 

6.041 

6. 107 

6.174 

6.243 

6.314 

9 

81 

6.314 

6.386 

6.460 

6.53s 

6.612 

6.691 

6.772 

6.85s 

6.940 

7.026 

7. IIS 

8 

82 

7. "5 

7-207 

7-300 

7-396 

7-495 

7  596 

7  -  700 

7.806 

7.916 

8.028 

8.144 

7 

83 

8.144 

8.264 

8.386 

8.513 

8.643 

8.777 

8.915 

9.058 

9-205 

9-357 

9-514 

6 

84 

9.514 

9.677 

9-845 

10.02 

10.20 

10.39 

10.58 

10.78 

10.99 

11.20 

11-43 

5 

86 

11-43 

11.66 

II  .91 

12.16 

12.43 

12.71 

13.00 

13  30 

13.62 

13-95 

14-30 

4 

86 

14-30 

14.67 

15.06 

15.46 

15-89 

16.35 

16.83 

17  34 

17.89 

18.46 

19.08 

3 

87 

19.08 

19  74 

20.45 

21.20 

22.02 

22.90 

23-86 

24.90 

26.03 

27.27 

28.64 

2 

88 

28.64 

30.14 

31.82 

33.69 

35-80 

38.19 

40.92 

44-07 

47.74 

52.08 

57-29 

1 

89 

57-29 

63.66 

71.62 

81.85 

95-49 

1x4.6 

143-2 

191. 0 

286.  5 

573.0 

00 

0° 

90° 

CO 

°.9 

°.8 

°.7 

°.6 

°.6 

°.4 

°.3 

°.2 

°.l 

°.0 

Angle 

bo  • 

<-3 

COTANGENTS,  angles  0°  to  45°.     Example,  cot  33  3°  =  1  5224 


382 


NATURAL  SECANTS 


SECANTS,  angles  0°  to  45°.    Example,  sec  33  3°  =  1.1964 


Angle 

°.o 

°.l 

°.2 

°.3 

°.4 

°.6 

°.6 

°.7 

°.8 

°.9 

<-3 

I . 0000 

90° 

0° 

I.OOOO 

0000 

0000 

0000 

0000 

0000 

0001 

0001 

0001 

0001 

0002 

89 

0 

1 

0002 

0002 

0002 

0003 

0003 

0003 

0004 

0004 

0005 

0006 

0006 

88 

0 

2 

ooo6 

0007 

OC07 

0008 

0009 

0010 

0010 

001 1 

0012 

0013 

0014 

87 

I 

3 

0014 

oois 

0016 

0017 

0018 

0019 

0020 

0021 

0022 

0023 

0024 

86 

I 

4 

0024 

0026 

0027 

0028 

0030 

0031 

0032 

0034 

003  5 

0037 

I .0038 

85 

I 

5 

I . 0038 

0040 

0041 

0043 

004s 

0046 

0048 

0050 

0051 

0053 

0055 

84 

2 

6 

00s  5 

00s  7 

OOS9 

0061 

0063 

0065 

0067 

0069 

0071 

0073 

0075 

83 

2 

7 

0075 

0077 

0079 

0082 

0084 

0086 

00S9 

0091 

0093 

0096 

0098 

82 

2 

8 

0098 

OIOI 

0103 

0106 

0108 

OIII 

0114 

01 16 

0119 

0122 

0125 

81 

3 

9 

0125 

0127 

0130 

0133 

0136 

0139 

0142 

0145 

0148 

0151 

I. 0154 

80° 

3 

10° 

I.OIS4 

OIS7 

0161 

0164 

0167 

0170 

0174 

0177 

0180 

0184 

0187 

79 

3 

11 

0187 

0191 

0194 

0198 

0201 

0205 

0209 

0212 

0216 

0220 

0223 

78 

4 

12 

0223 

0227 

0231 

023s 

0239 

0243 

0247 

0251 

0255 

0259 

0263 

77 

4 

13 

0263 

0267 

0271 

0276 

0280 

0284 

0288 

0293 

0297 

0302 

0306 

76 

4 

14 

0306 

03 1 1 

031S 

0320 

0324 

0329 

0334 

0338 

0343 

0348 

I  0353 

75 

5 

15 

I -0353 

0358 

0363 

0367 

0372 

0377 

0382 

0388 

0393 

0398 

0403 

74 

5 

16 

0403 

0408 

0413 

0419 

0424 

0429 

0435 

0440 

0446 

0451 

04S7 

73 

5 

17 

04s  7 

0463 

0468 

0474 

0480 

0485 

0491 

0497 

0503 

0509 

0515 

72 

6 

18 

0515 

0521 

0527 

0533 

0539 

0545 

0551 

0557 

0564 

0570 

0576 

71 

6 

19 

0576 

0583 

0589 

059s 

0602 

0608 

0615 

0622 

0628 

0635 

1.0642 

70° 

7 

20° 

1.0642 

0649 

0655 

0662 

0669 

0676 

0683 

0690 

0697 

0704 

071 1 

69 

7 

21 

071 1 

0719 

0726 

0733 

0740 

0748 

0755 

0763 

0770 

0778 

0785 

68 

7 

22 

078s 

0793 

0801 

0808 

0816 

0824 

0832 

0840 

0848 

0856 

0864 

67 

8 

23 

0864 

0872 

0880 

0888 

0896 

0904 

0913 

0921 

0929 

0938 

0946 

66 

8 

24 

0946 

095s 

0963 

0972 

0981 

0989 

0998 

1007 

1016 

1025 

I . 1034 

65 

9 

26 

I. 1034 

1043 

1052 

106 1 

1070 

1079 

1089 

1098 

1 107 

1117 

1126 

64 

9 

26 

1 1 26 

1 136 

1 145 

iiSS 

1 164 

1174 

1 184 

1 194 

1203 

1213 

1223 

63 

10 

27 

1223 

1233 

1243 

1253 

1264 

1274 

1284 

1294 

1305 

1215 

1326 

62 

10 

28 

1326 

1336 

1347 

1357 

1368 

1379 

1390 

1401 

1412 

1423 

1434 

61 

II 

29 

1434 

1445 

1456 

1467 

1478 

1490 

1501 

1512 

1524 

1 535 

I -1547 

60° 

II 

30° 

I -1547 

ISS9 

1570 

1582 

1594 

1606 

1618 

1630 

1642 

1654 

1666 

59 

12 

31 

1666 

1679 

1691 

1703 

1716 

1728 

1741 

1753 

1766 

1779 

1792 

58 

13 

32 

1792 

1805 

1818 

1831 

1844 

1857 

1870 

1883 

1897 

1910 

1924 

67 

13 

33 

1924 

1937 

1951 

1964 

1978 

1992 

2006 

2020 

2034 

2048 

2062 

66 

14 

34 

2062 

2076 

2091 

2105 

2120 

2134 

2149 

2163 

2178 

2193 

1.2208 

55 

15 

35 

1.2208 

2223 

2238 

2253 

2268 

2283 

2299 

2314 

2329 

2345 

2361 

54 

15 

36 

2361 

2376 

2392 

2408 

2424 

2440 

2456 

2472 

2489 

2505 

2521 

53 

16 

37 

2521 

2538 

2554 

2571 

2588 

2605 

2622 

2639 

2656 

2673 

2690 

52 

17 

38 

2690 

2708 

2725 

2742 

2760 

2778 

2796 

2813 

2831 

2849 

2868 

51 

18 

39 

2868 

2886 

2904 

2923 

2941 

2960 

2978 

2997 

3016 

3035 

1-3054 

60° 

19 

40° 

I -3054 

3073 

3093 

3112 

3131 

3151 

3171 

3190 

3210 

3230 

3250 

49 

20 

41 

3250 

3270 

3291 

3311 

3331 

3352 

3373 

3393 

3414 

3435 

3456 

48 

21 

42 

3456 

3478 

3499 

3520 

3542 

3563 

3585 

3607 

3629 

3651 

3673 

47 

22 

43 

3673 

3696 

3718 

3741 

3763 

3786 

3809 

3832 

385s 

3878 

3902 

46 

23 

44 

3902 

3925 

3949 

3972 

3996 

4020 

4044 

4069 

4093 

4118 

I. 4142 

45° 

24 

45° 

1-4142 

°.9 

°.8 

°.7 

°.6 

°.5 

°.4 

°.3 

°.2 

°.l 

°.o 

Angle 

COSECANTS,  angles  45°  to  90°.    Example,  esc  66. 6°  =  1.0896 


AND  COSECANTS 


383 


SECANTS,  angles  45°  to  90°.     Example,  sec  66.6°  =  2  518 


Angle 

°.o 

°.l 

°.2 

°.3 

°.4 

°.6 

°.6 

°.7 

°.8 

°.9 

I. 4142 

45° 

46° 

I. 4142 

4167 

4192 

4217 

4242 

4267 

4293 

4318 

4344 

4370 

4396 

44 

25 

46 

4396 

4422 

4448 

4474 

4501 

4527 

4554 

4581 

4608 

4635 

4663 

43 

27 

47 

4663 

4690 

4718 

4746 

4774 

4802 

4830 

4859 

4887 

4916 

4945 

42 

28 

48 

4945 

4974 

5003 

5032 

5062 

5092 

5121 

S151 

5182 

5212 

5243 

41 

30 

49 

5243 

5273 

5304 

5335 

5366 

5398 

5429 

5461 

5493 

5525 

I.S557 

40 

31 

50° 

I-S557 

5590 

5622 

565s 

5688 

5721 

5755 

5788 

5822 

5856 

5890 

39 

33 

51 

5890 

5925 

5959 

5994 

6029 

6064 

6099 

6135 

6171 

6207 

6243 

38 

35 

62 

6243 

6279 

6316 

6353 

6390 

6427 

6464 

6502 

6540 

6578 

6616 

37 

37 

53 

6616 

6655 

6694 

6733 

6772 

6812 

6852 

6892 

6932 

6972 

7013 

36 

40 

54 

7013 

7054 

709s 

7137 

7179 

7221 

7263 

7305 

7348 

7391 

I  -  7434 

36 

42 

65 

1-7434 

7478 

7522 

7566 

7610 

7655 

7700 

7745 

7791 

7837 

7883 

34 

45 

56 

7883 

7929 

7976 

8023 

8070 

8118 

8166 

8214 

8263 

8312 

8361 

33 

48 

67 

8361 

8410 

8460 

8510 

8561 

8612 

8663 

8714 

8766 

8818 

8871 

32 

51 

68 

8871 

8924 

8977 

9031 

9084 

9139 

9194 

9249 

9304 

9360 

I .9416 

31 

54 

69 

I. 9416 

9473 

9530 

9587 

9645 

9703 

9762 

9821 

9880 

9940 

2 . 0000 

30° 

58 

60° 

2.000 

2.006 

2  .012 

2.018 

2.025 

2.031 

2.037 

2.043 

2.050 

2.056 

2.063 

29 

6 

61 

2.063 

2.069 

2.076 

2.082 

2.089 

2.096 

2.103 

2.109 

2. 116 

2.123 

2.130 

28 

7 

62 

2.130 

2.137 

2.144 

2. 151 

2.158 

2. 166 

2.173 

2.180 

2.188 

2.195 

2.203 

27 

7 

63 

2.203 

2.210 

2.218 

2.226 

2-233 

2. 241 

2-249 

2.257 

2.265 

2.273 

2.281 

26 

8 

64 

2.281 

2.289 

2.298 

2.306 

2.314 

2.323 

2.331 

2.340 

2-349 

2-357 

2.366 

25 

8 

66 

2.366 

2-375 

2-384 

2.393 

2  .402 

2. 411 

2.421 

2.430 

2-439 

2-449 

2-459 

24 

9 

66 

2-459 

2.468 

2.478 

2.488 

2.498 

2.508 

2.518 

2.528 

2-538 

2.549 

2-559 

23 

10 

67 

2-559 

2.570 

2.581 

2.591 

2.602 

2.613 

2.624 

2 .  635 

2-647 

2.658 

2.669 

22 

II 

68 

2.669 

2.681 

2.693 

2.705 

2.716 

2.729 

2.741 

2.753 

2.765 

2-778 

2.790 

21 

12 

69 

2-790 

2.803 

2.816 

2.829 

2.842 

2.855 

2.869 

2.882 

2.896 

2  .910 

2.924 

20° 

13 

70° 

2-924 

2-938 

2.952 

2.967 

2.981 

2.996 

3.001 

3.026 

3-041 

3-056 

3.072 

19 

IS 

71 

3.072 

3-087 

3 -103 

3-119 

3-135 

3-152 

3.168 

3.185 

3.202 

3.219 

3-236 

18 

16 

72 

3-236 

3-254 

3-271 

3.289 

3.307 

3-326 

3 .  344 

3.363 

3-382 

3-401 

3-420 

17 

18 

73 

3.420 

3-440 

3.460 

3.480 

3.500 

3-521 

3-542 

3.563 

3-584 

3.606 

3.628 

16 

21 

74 

3.628 

3-650 

3-673 

3-695 

3.719 

3-742 

3.766 

3.790 

3-814 

3-839 

3.864 

16 

24 

76 

3-864 

3.889 

3-915 

3-941 

3.967 

3-994 

4.021 

4.049 

4.077 

4-105 

4-134 

14 

27 

76 

4-^34 

4.163 

4-192 

4.222 

4.253 

4.284 

4-315 

4-347 

4-379 

4.412 

4-445 

13 

31 

77 

4-445 

4-479 

4-514 

4-549 

4.584 

4.620 

4.657 

4.694 

4-732 

4-771 

4-810 

12 

36 

78 

4.810 

4.850 

4.890 

4-931 

4.973 

5.016 

5 -059 

5- 103 

5-148 

5-194 

5-241 

11 

43 

79 

5-241 

5-288 

5-337 

5.386 

5. 436 

5. 487 

5.540 

5-593 

5-647 

5.702 

5-759 

10 

52 

80° 

5-759 

S-816 

5-875 

5 -935 

5.996 

6.059 

6.123 

6.188 

6-255 

6.323 

6.392 

9 

81 

6-392 

6.464 

6-537 

6. 611 

.6.687 

6.765 

6.845 

6.927 

7  .011 

7.097 

7-185 

8 

82 

7-185 

7.276 

7-368 

7-463 

7.561 

7.661 

7.764 

7-870 

7-979 

8.091 

8 .206 

7 

83 

8-206 

8.324 

8.446 

8-571 

8.700 

8.834 

8.971 

9113 

9-259 

9. 411 

9-567 

6 

84 

9-567 

9.728 

9-895 

10.07 

10.25 

10.43 

10.63 

10.83 

11-03 

11.25 

11-47 

6 

85 

II -47 

II. 71 

11-95 

12 .  20 

12.47 

12.75 

13.03 

13-34 

13.65 

13.99 

14-34 

4 

86 

14-34 

14.70 

15-09 

1550 

15. 93 

16.38 

16.86 

17-37 

17.91 

18.49 

i9- 11 
28.65 

3 

87 

I9-II 

19.77 

20.47 

21.23 

22.04 

22.93 

23.88 

24.92 

26.05 

27.29 

2 

88 

28-65 

30.16 

31-84 

33-71 

35.81 

38.20 

40.93 

44.08 

47.75 

52.09 

57-30 

1 

89 
90° 

57-30 

00 

63-66 

71 .62 

81-35 

95-49 

114. 6 

143.2 

191 .0 

286.5 

573.0 

00 

0 

°.9 

°.8 

°.7 

°.6 

°.5 

°.4 

°.3 

°.2 

°.l 

°.o 

Angle 

6 

<:-5 

COSECANTS,  angles  0°  to  45°.     Example,  esc  33  3°  =  1.8214 


INDEX 

(Numbers  refer  to  pages.) 


Acceleration,  121,  144. 

angular,  177. 

components,  150 

graphs,  128. 
Action  and  reaction,  43. 
Amplitude  of  simple  harmonic  motion,  131. 
Analysis  of  a  truss,  57. 
Angle  of  repose,  75. 
Angular  acceleration,  177. 

impulse,  237. 

momentum,  237. 

velocity,  176,  281. 
Anti-resultant,  7. 
Arm  of  a  couple,  18. 
Atwood's  machine,  139. 

Band  brake,  85. 
Belt  friction,  83. 
Blow,  232. 

Cables,  103. 
Catenary,  107. 
Center  of  gravity,  87. 

experimental  determination,  8g 

motion  of,  155. 
Center  of  gyration,  169. 
Center  of  percussion,  241. 
Centroid,  86,  90,  98. 
Coefficient  of  friction,  75,  222. 
Collision,  232. 
Composition  of 

angular  velocities,  281. 

couples,  29. 

forces,  7. 
Conical  pendulum,  161. 
Conservation  of 

angular  momentum,  240. 

energy,  300. 
Couples,  18,  28. 
Cranes,  analysis  of,  64. 

D'Alembert's  principle,  300. 
Dimensions  of  units,  302. 


Dynamometers,  198,  199, 
Dyne,  143. 

Efficiency,  211,  212. 
Elevation  of  outer  rail,  161, 
Energy,  193,  195. 
Equilibrant,  7. 
Equilibrium,  34. 
Erg,  190. 

Force 

characteristics,  4. 

definition,  4. 

external,  34,  158. 

internal,  34,  158. 

line  of  action  of,  5. 

moment  of,  16. 

polygon,  12. 

sense  of,  5. 

tractive,  76. 

transmissibility  of,  6. 
Forces 

cohnear,  7. 

composition  of,  7. 

concurrent,  7. 

coplanar,  7. 

parallelogram  of,  7. 

parallelopiped  of,  13. 

resolution  of,  7. 

triangle  of,  7. 
Force  of  inertia,  157. 
Frameworks,  54,  64. 
Free  axes,  294. 
Frequency 

of  simple  harmonic  motion,  131. 
Friction,  74,  221. 
Friction  of 

belts,  84. 

inclined  planes,  78. 

journals,  82,  225. 

pivots,  223. 

screws,  81. 

wedges,  79. 
Friction,  rolling,  268. 


38s 


386 


Geepound,  143. 
Graphs,  126. 

Gravity,  acceleration  of,  5. 
Gyration,  center  of,  169. 

radius  of,  169,  292,  310. 
Gyro-compass,  247. 
Gyroscope,  243. 
Gyro-stabilizer,  247,  248. 
Gyrostat,  243,  288. 
Gyrostatic  reaction,  253. 

Harmonic  motion,  131. 
Hodograph,  145. 
Hoists,  212. 
Horse-power,  197. 
Horse-power-hour,  190. 

Impact,  232. 
Impulse,  angular,  242. 

linear,  231. 
Inclined  plane,  78,  214. 
Indicator,  200. 
Indicator  card,  201. 
Inertia,  156. 

circle,  320. 

ellipse,  321. 

moment  of,  168,  292,  308. 

product  of,  293,  316. 
Input,  211. 
Instantaneous  axis,  281. 

center,  260. 

Joule,  190. 
Journal  friction,  82. 

Kilogram,  4. 
Kinetic  Energy,  193. 

Lami's  theorem,  39. 
Laws  of  motion,  155. 
Lever,  i. 
Locomotive: 

cylinder  effort,  206. 

side  rods,  166. 

Mass,  142. 
Mass-center,  158. 
Moment 

center  of,  16. 

moment-sum,  17. 

of  momentum,  237 

origin  of,  16. 

statical,  91. 


Moment  of 

a  body,  87. 

a  couple,  18. 

a  force,  16,  27. 

a  line,  91. 

a  solid,  91. 

a  surface,  91. 
Moment  of  inertia,  168,  292,  308. 

graphical  determination,  320. 

inclined  axis  theorem,  318. 

parallel  axes  theorem,  171,  311. 

principal  axes  of,  292,  318. 

rectangular  axes  theorem,  313. 
Moments,  principle  of,  17,  28. 
Momentum,  angular,  237. 

linear,  228. 

moment  of,  237. 
Mono-rail  car,  247. 
Motion 

curvilinear,  144. 

graphs,  126. 

laws  of,  155. 

non-uniform,  118. 

precessional,  245. 

plane,  256. 

rectilinear,  118. 

relative,  273. 

rotational,  176. 

simple  harmonic,  131. 

solid,  280. 

spherical,  280. 

translational,  163. 

uniform,  118. 

imiformly  accelerated,  121,  125. 
Motion  of 

center  of  gravity,  155. 

of  rotation,  176. 

of  translation,  163. 

Newton's  laws,  2,  155. 

Outer  rail,  elevation  of,  161. 
Output,  211. 

Parallelogram  of  forces,  7. 
Parallelepiped  of  forces,  13. 
Parallel  axes  theorem,  171,  311. 
Particle,  155. 
Pendulum,  gravity,  182. 

torsional,  187. 
Percussion,  center  of. 
Period  of  simple  harmonic  motion,  131. 
Pole,  22. 


387 


Polygon  equilibrium,  22. 

funicular,  22. 

of  forces,  12. 

string,  22. 
Potential  energy,  195. 
Pound,  4. 
Power,  196. 

measurement  of,  198. 

of  a  locomotive,  200. 
Precession,  245,  288. 
Principal  axes,  000,  000. 
Principles  of 

dynamics,  297. 

moments,  17,  28. 

work  and  energy,  203. 
Problems,  statically  indeterminate,  43. 
Product  of  inertia,  293,  316. 
Projectile,  153. 
Prony  brake,  198. 
Pulley,  69,  217. 

Radius  of  gyration,  169,  310. 

parallel  axis  theorem,  171. 
Rate  of  a 

scalar  quantity,  123. 

vector  quantity,  148. 
Rays,  22. 

Rectangular  axes  theorem. 
Relative  motion,  273. 
Repose,  angle  of,  5. 
Resolution  of 

acceleration,  149. 

couples,  30. 

forces,  7. 

velocity,  150,  000. 
Resultant,  7. 
Rolling  resistance,  268. 
Rotation,  176. 


Speed,  144. 

Spherical  motion,  54 

Statical  moment,  91. 

Statically  indeterminate  problems,  45,  59. 

Stress,  55. 

Stress  diagrams,  59. 

String  polygon, 


22. 


Tackle,  218. 
Tension  in  a  cord,  42. 
Theorem 

inclined  axis, 

Lami's,  39. 

Pappus',  96. 

parallel  axes,  171,  311. 

rectangular  axes,  313. 
Torpedo,  246. 
Torque,  16. 

Torsion  pendulum,  187. 
Train  resistance,  208. 
Translation,  163. 
Triangle  of  forces,  7. 
Truss,  analysis  of,  57,  59. 

Units 
absolute,  138. 
dimensions  of,  302. 
fundamental,  302. 
gravitation,  4. 
systematic,  143. 

Vector  diagrams,  6. 
Vector  quantity,  5. 

rate  of,  148. 
Velocity,  118,  144. 

angular,  176. 

components  of,  148. 

graphs,  127,  128. 


Scalar  quantity,  5. 

rate  of  a,  123. 
Screw,  81,  216. 

Simple  harmonic  motion,  131. 
Slug,  143- 
Space  diagram,  6. 


Watt,  197. 
Watt-hour,  190. 
Wedge,  79,  215. 
Work,  189. 
Work  and  energy, 
principle  of,  203. 
(Numbers  refer  to  pages.) 


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